zoukankan html css js c++ java

最小割

复习一波最小割...

理论知识：

参考资料。

BZOJ1934 善意的投票

题意：n个人分别喜欢睡/不睡午觉。改变一个人的喜好要花费1代价。有m对朋友，每对朋友的喜好不同会产生1代价。求最小代价。

解：经典题了.....

每个人向s，t连边，求最小割。如果割掉某个边就代表选哪边。朋友之间连双向边，代表如果选的不同的话还要割一刀。

考虑为什么不会有点两条边都被割。因为两条边都被割表明有个跟它相连的没割上面，又有个跟它相连的没割下面，而这显然是不可能的，因为它们会通过这个点间接的出现一条流。

 1 #include <bits/stdc++.h>
 2 
 3 const int N = 3010, INF = 0x3f3f3f3f;
 4 
 5 struct Edge {
 6     int nex, v, c;
 7     Edge(int NEX = 0, int V = 0, int C = 0) {
 8         nex = NEX;
 9         v = V;
10         c = C;
11     }
12 }edge[2000010]; int tp = 1;
13 
14 int e[N], d[N], vis[N], Time, cur[N], cnt;
15 std::queue<int> Q;
16 int n, a[N];
17 
18 inline void add(int x, int y, int z) {
19     edge[++tp] = Edge(e[x], y, z);
20     e[x] = tp;
21     edge[++tp] = Edge(e[y], x, 0);
22     e[y] = tp;
23     return;
24 }
25 
26 inline bool BFS(int s, int t) {
27     Time++;
28     vis[s] = Time;
29     d[s] = 0;
30     Q.push(s);
31     while(Q.size()) {
32         int x = Q.front();
33         Q.pop();
34         for(int i = e[x]; i; i = edge[i].nex) {
35             int y = edge[i].v;
36             if(vis[y] == Time || !edge[i].c) continue;
37             vis[y] = Time;
38             d[y] = d[x] + 1;
39             Q.push(y);
40         }
41     }
42     return vis[t] == Time;
43 }
44 
45 int DFS(int x, int t, int maxF) {
46     if(x == t) return maxF;
47     int ans = 0;
48     for(int i = cur[x]; i; i = edge[i].nex) {
49         int y = edge[i].v;
50         if(d[y] != d[x] + 1 || !edge[i].c) continue;
51         int temp = DFS(y, t, std::min(edge[i].c, maxF - ans));
52         if(!temp) d[y] = INF;
53         ans += temp;
54         edge[i].c -= temp;
55         edge[i ^ 1].c += temp;
56         if(ans == maxF) break;
57         cur[x] = i;
58     }
59     return ans;
60 }
61 
62 inline int solve(int s, int t) {
63     int ans = 0;
64     while(BFS(s, t)) {
65         memcpy(cur + 1, e + 1, cnt * sizeof(int));
66         ans += DFS(s, t, INF);
67     }
68     return ans;
69 }
70 
71 int main() {
72     int m;
73     scanf("%d%d", &n, &m);
74     int s = n + 1, t = n + 2;
75     cnt = t;
76     for(int i = 1; i <= n; i++) {
77         scanf("%d", &a[i]);
78         if(a[i]) {
79             add(s, i, 1);
80         }
81         else {
82             add(i, t, 1);
83         }
84     }
85     for(int i = 1, x, y; i <= m; i++) {
86         scanf("%d%d", &x, &y);
87         add(x, y, 1);
88         add(y, x, 1);
89     }
90     int ans = solve(s, t);
91     printf("%d
", ans);
92     return 0;
93 }

AC代码

BZOJ2127 happiness

题意：n个人，m对朋友。每个人选文科，理科都分别有贡献，每对朋友同时选文/选理也有贡献。求最大贡献。

解：先设每个人同时选了文理。考虑割掉。

s向人连选文的贡献，人向t连选理的。一对朋友之间，考虑到如果有任一个选了文，就拿不到全选理的贡献，所以建图大概长这样...

然后求最小割，拿总贡献减去，就是最大贡献了。

  1 #include <bits/stdc++.h>
  2 
  3 const int N = 100010, INF = 0x3f3f3f3f;
  4 
  5 struct Edge {
  6     int nex, v, c;
  7     Edge(int NEX = 0, int V = 0, int C = 0) {
  8         nex = NEX;
  9         v = V;
 10         c = C;
 11     }
 12 }edge[2000010]; int tp = 1;
 13 
 14 int e[N], d[N], vis[N], Time, cur[N], cnt;
 15 std::queue<int> Q;
 16 int n, m, s, t;
 17 
 18 inline void add(int x, int y, int z) {
 19     edge[++tp] = Edge(e[x], y, z);
 20     e[x] = tp;
 21     edge[++tp] = Edge(e[y], x, 0);
 22     e[y] = tp;
 23     return;
 24 }
 25 
 26 inline bool BFS(int s, int t) {
 27     Time++;
 28     vis[s] = Time;
 29     d[s] = 0;
 30     Q.push(s);
 31     while(Q.size()) {
 32         int x = Q.front();
 33         Q.pop();
 34         for(int i = e[x]; i; i = edge[i].nex) {
 35             int y = edge[i].v;
 36             if(vis[y] == Time || !edge[i].c) continue;
 37             vis[y] = Time;
 38             d[y] = d[x] + 1;
 39             Q.push(y);
 40         }
 41     }
 42     return vis[t] == Time;
 43 }
 44 
 45 int DFS(int x, int t, int maxF) {
 46     if(x == t) return maxF;
 47     int ans = 0;
 48     for(int i = cur[x]; i; i = edge[i].nex) {
 49         int y = edge[i].v;
 50         if(d[y] != d[x] + 1 || !edge[i].c) continue;
 51         int temp = DFS(y, t, std::min(edge[i].c, maxF - ans));
 52         if(!temp) d[y] = INF;
 53         ans += temp;
 54         edge[i].c -= temp;
 55         edge[i ^ 1].c += temp;
 56         if(ans == maxF) break;
 57         cur[x] = i;
 58     }
 59     return ans;
 60 }
 61 
 62 inline int solve(int s, int t) {
 63     int ans = 0;
 64     while(BFS(s, t)) {
 65         memcpy(cur + 1, e + 1, cnt * sizeof(int));
 66         ans += DFS(s, t, INF);
 67     }
 68     return ans;
 69 }
 70 
 71 inline int id(int i, int j) {
 72     return (i - 1) * m + j;
 73 }
 74 
 75 int main() {
 76     scanf("%d%d", &n, &m);
 77     s = n * m + 1, t = s + 1;
 78     int ans = 0, x;
 79     cnt = t;
 80     for(int i = 1; i <= n; i++) {
 81         for(int j = 1; j <= m; j++) {
 82             scanf("%d", &x);
 83             ans += x;
 84             add(s, id(i, j), x);
 85         }
 86     }
 87     for(int i = 1; i <= n; i++) {
 88         for(int j = 1; j <= m; j++) {
 89             scanf("%d", &x);
 90             ans += x;
 91             add(id(i, j), t, x);
 92         }
 93     }
 94     for(int i = 1; i < n; i++) {
 95         for(int j = 1; j <= m; j++) {
 96             scanf("%d", &x);
 97             ans += x;
 98             add(s, ++cnt, x);
 99             add(cnt, id(i, j), INF);
100             add(cnt, id(i + 1, j), INF);
101         }
102     }
103     for(int i = 1; i < n; i++) {
104         for(int j = 1; j <= m; j++) {
105             scanf("%d", &x);
106             ans += x;
107             add(++cnt, t, x);
108             add(id(i, j), cnt, INF);
109             add(id(i + 1, j), cnt, INF);
110         }
111     }
112     for(int i = 1; i <= n; i++) {
113         for(int j = 1; j < m; j++) {
114             scanf("%d", &x);
115             ans += x;
116             add(s, ++cnt, x);
117             add(cnt, id(i, j), INF);
118             add(cnt, id(i, j + 1), INF);
119         }
120     }
121     for(int i = 1; i <= n; i++) {
122         for(int j = 1; j < m; j++) {
123             scanf("%d", &x);
124             ans += x;
125             add(++cnt, t, x);
126             add(id(i, j), cnt, INF);
127             add(id(i, j + 1), cnt, INF);
128         }
129     }
130     ans -= solve(s, t);
131     printf("%d
", ans);
132     return 0;
133 }

AC代码

注意这题建图画风跟别的题不一样，是因为按照之前的建图，当一对朋友同时选0的时候，仍要损失同时选1的代价，而这部分是我们难以计算的。

BZOJ1976 能量魔方

题意：有n³个块构成了立方体，有些位置填了0/1，别的地方自己填。6连通的块之间如果颜色不同就会有1贡献，求最大贡献。

解：前两题都是最大化相同，最小化不同，这题反其道而行之，何如？

考虑到这些格子和6连通关系构成了二分图。我们把其中一个部分全部反色，不就成了最大化相同吗？直接套用第一题的做法。

  1 #include <bits/stdc++.h>
  2 
  3 const int N = 100010, INF = 0x3f3f3f3f;
  4 
  5 struct Edge {
  6     int nex, v, c;
  7     Edge(int NEX = 0, int V = 0, int C = 0) {
  8         nex = NEX;
  9         v = V;
 10         c = C;
 11     }
 12 }edge[2000010]; int tp = 1;
 13 
 14 int e[N], d[N], vis[N], Time, cur[N], cnt;
 15 std::queue<int> Q;
 16 int n;
 17 
 18 inline void add(int x, int y, int z) {
 19     edge[++tp] = Edge(e[x], y, z);
 20     e[x] = tp;
 21     edge[++tp] = Edge(e[y], x, 0);
 22     e[y] = tp;
 23     return;
 24 }
 25 
 26 inline bool BFS(int s, int t) {
 27     Time++;
 28     vis[s] = Time;
 29     d[s] = 0;
 30     Q.push(s);
 31     while(Q.size()) {
 32         int x = Q.front();
 33         Q.pop();
 34         for(int i = e[x]; i; i = edge[i].nex) {
 35             int y = edge[i].v;
 36             if(vis[y] == Time || !edge[i].c) continue;
 37             vis[y] = Time;
 38             d[y] = d[x] + 1;
 39             Q.push(y);
 40         }
 41     }
 42     return vis[t] == Time;
 43 }
 44 
 45 int DFS(int x, int t, int maxF) {
 46     if(x == t) return maxF;
 47     int ans = 0;
 48     for(int i = cur[x]; i; i = edge[i].nex) {
 49         int y = edge[i].v;
 50         if(d[y] != d[x] + 1 || !edge[i].c) continue;
 51         int temp = DFS(y, t, std::min(edge[i].c, maxF - ans));
 52         if(!temp) d[y] = INF;
 53         ans += temp;
 54         edge[i].c -= temp;
 55         edge[i ^ 1].c += temp;
 56         if(ans == maxF) break;
 57         cur[x] = i;
 58     }
 59     return ans;
 60 }
 61 
 62 inline int solve(int s, int t) {
 63     int ans = 0;
 64     while(BFS(s, t)) {
 65         memcpy(cur + 1, e + 1, cnt * sizeof(int));
 66         ans += DFS(s, t, INF);
 67     }
 68     return ans;
 69 }
 70 
 71 inline int id(int x, int y, int z) {
 72     return (x - 1) * n * n + (y - 1) * n + z;
 73 }
 74 
 75 char str[N];
 76 
 77 int main() {
 78     scanf("%d", &n);
 79     int s = n * n * n + 1, t = s + 1, ans = 0;
 80     cnt = t;
 81     for(int i = 1; i <= n; i++) {
 82         for(int j = 1; j <= n; j++) {
 83             scanf("%s", str + 1);
 84             for(int k = 1; k <= n; k++) {
 85                 if(str[k] == '?');
 86                 else if((str[k] == 'P') == ((i + j + k) & 1)) {
 87                     add(s, id(i, j, k), INF);
 88                 }
 89                 else {
 90                     add(id(i, j, k), t, INF);
 91                 }
 92 
 93                 if(i > 1) {
 94                     add(id(i, j, k), id(i - 1, j, k), 1);
 95                     add(id(i - 1, j, k), id(i, j, k), 1);
 96                     ans++;
 97                 }
 98                 if(j > 1) {
 99                     add(id(i, j, k), id(i, j - 1, k), 1);
100                     add(id(i, j - 1, k), id(i, j, k), 1);
101                     ans++;
102                 }
103                 if(k > 1) {
104                     add(id(i, j, k), id(i, j, k - 1), 1);
105                     add(id(i, j, k - 1), id(i, j, k), 1);
106                     ans++;
107                 }
108             }
109         }
110     }
111     printf("%d
", ans - solve(s, t));
112     return 0;
113 }

AC代码

BZOJ2132 圈地计划

题意：给你一个矩阵，每个位置放0有贡献，放1有贡献，4连通的格子放不同的还有贡献，求最大贡献。

解：类似上一题，黑白染色之后把黑色格子放的全部反转，就有相同相邻的格子会产生贡献。注意01分别的贡献也要相应的反转。

然后使用上一题的做法即可。

  1 #include <bits/stdc++.h>
  2 
  3 const int N = 100010, INF = 0x3f3f3f3f;
  4 
  5 struct Edge {
  6     int nex, v, c;
  7     Edge(int NEX = 0, int V = 0, int C = 0) {
  8         nex = NEX;
  9         v = V;
 10         c = C;
 11     }
 12 }edge[2000010]; int tp = 1;
 13 
 14 int e[N], d[N], vis[N], Time, cur[N], cnt;
 15 std::queue<int> Q;
 16 int n, m, a[101][101], b[101][101], c[101][101];
 17 
 18 inline void add(int x, int y, int z) {
 19     edge[++tp] = Edge(e[x], y, z);
 20     e[x] = tp;
 21     edge[++tp] = Edge(e[y], x, 0);
 22     e[y] = tp;
 23     return;
 24 }
 25 
 26 inline bool BFS(int s, int t) {
 27     Time++;
 28     vis[s] = Time;
 29     d[s] = 0;
 30     Q.push(s);
 31     while(Q.size()) {
 32         int x = Q.front();
 33         Q.pop();
 34         for(int i = e[x]; i; i = edge[i].nex) {
 35             int y = edge[i].v;
 36             if(vis[y] == Time || !edge[i].c) continue;
 37             vis[y] = Time;
 38             d[y] = d[x] + 1;
 39             Q.push(y);
 40         }
 41     }
 42     return vis[t] == Time;
 43 }
 44 
 45 int DFS(int x, int t, int maxF) {
 46     if(x == t) return maxF;
 47     int ans = 0;
 48     for(int i = cur[x]; i; i = edge[i].nex) {
 49         int y = edge[i].v;
 50         if(d[y] != d[x] + 1 || !edge[i].c) continue;
 51         int temp = DFS(y, t, std::min(edge[i].c, maxF - ans));
 52         if(!temp) d[y] = INF;
 53         ans += temp;
 54         edge[i].c -= temp;
 55         edge[i ^ 1].c += temp;
 56         if(ans == maxF) break;
 57         cur[x] = i;
 58     }
 59     return ans;
 60 }
 61 
 62 inline int solve(int s, int t) {
 63     int ans = 0;
 64     while(BFS(s, t)) {
 65         memcpy(cur + 1, e + 1, cnt * sizeof(int));
 66         ans += DFS(s, t, INF);
 67     }
 68     return ans;
 69 }
 70 
 71 inline int id(int i, int j) {
 72     return (i - 1) * m + j;
 73 }
 74 
 75 int main() {
 76     scanf("%d%d", &n, &m);
 77     int s = n * m + 1, t = s + 1, ans = 0;
 78     cnt = t;
 79     for(int i = 1; i <= n; i++) {
 80         for(int j = 1; j <= m; j++) {
 81             scanf("%d", &a[i][j]);
 82             ans += a[i][j];
 83         }
 84     }
 85     for(int i = 1; i <= n; i++) {
 86         for(int j = 1; j <= m; j++) {
 87             scanf("%d", &b[i][j]);
 88             ans += b[i][j];
 89             if((i + j) & 1) {
 90                 std::swap(a[i][j], b[i][j]);
 91             }
 92             add(s, id(i, j), a[i][j]);
 93             add(id(i, j), t, b[i][j]);
 94         }
 95     }
 96     for(int i = 1; i <= n; i++) {
 97         for(int j = 1; j <= m; j++) {
 98             scanf("%d", &c[i][j]);
 99             if(i > 1) {
100                 add(id(i, j), id(i - 1, j), c[i][j] + c[i - 1][j]);
101                 add(id(i - 1, j), id(i, j), c[i][j] + c[i - 1][j]);
102                 ans += c[i][j] + c[i - 1][j];
103             }
104             if(j > 1) {
105                 add(id(i, j), id(i, j - 1), c[i][j] + c[i][j - 1]);
106                 add(id(i, j - 1), id(i, j), c[i][j] + c[i][j - 1]);
107                 ans += c[i][j] + c[i][j - 1];
108             }
109         }
110     }
111 
112     printf("%d
", ans - solve(s, t));
113 
114     return 0;
115 }

AC代码

LOJ#2146 寿司餐厅

题意：有个寿司序列，每个子串都有个价值，每个寿司有个种类。你可以无限次的取，每次取一个子串，那么就能得到这个子串的所有子串的贡献。但是每个子串[l, r]的贡献不管取了多少次都只会算一次。如果最终这n个寿司中你吃到过其中i(i > 0)个x种类的寿司，那么就要付出m * x * x + i * x的钱。求你能得到的最大(总价值 - 钱数)。

解：在那想50分状压怎么做，结果想着就把正解搞出来了，状压还是不会...

画出矩形来，发现每次就是取一个左下角的前缀和，相当于取了一个点就要取它左下的所有点。这是最大权闭合子图。然后怎么算付的钱呢？每个寿司(就是i，i那个位置)连到一个(-x)的点。然后所有(-x)的点再连到一个(-x²)的点即可。

  1 #include <bits/stdc++.h>
  2 
  3 const int N = 110, M = 100010, INF = 0x3f3f3f3f;
  4 
  5 struct Edge {
  6     int nex, v, c;
  7     Edge(int NEX = 0, int V = 0, int C = 0) {
  8         nex = NEX;
  9         v = V;
 10         c = C;
 11     }
 12 }edge[2000010]; int tp = 1;
 13 
 14 int e[M], a[N], d[N][N], n, m, val[M], dis[M], vis[M], Time, cnt, cur[M], id[N][N], use[1010];
 15 std::queue<int> Q;
 16 
 17 inline void add(int x, int y, int z) {
 18     //printf("add %d %d %d 
", x, y, z);
 19     edge[++tp] = Edge(e[x], y, z);
 20     e[x] = tp;
 21     edge[++tp] = Edge(e[y], x, 0);
 22     e[y] = tp;
 23     return;
 24 }
 25 
 26 inline bool BFS(int s, int t) {
 27     Time++;
 28     vis[s] = Time;
 29     dis[s] = 0;
 30     Q.push(s);
 31     while(Q.size()) {
 32         int x = Q.front();
 33         Q.pop();
 34         //printf("BFS x = %d 
", x);
 35         for(int i = e[x]; i; i = edge[i].nex) {
 36             int y = edge[i].v;
 37             if(vis[y] == Time || !edge[i].c) {
 38                 continue;
 39             }
 40             //printf("BFS %d -> %d 
", x, y);
 41             vis[y] = Time;
 42             dis[y] = dis[x] + 1;
 43             Q.push(y);
 44         }
 45     }
 46     return vis[t] == Time;
 47 }
 48 
 49 int DFS(int x, int t, int maxF) {
 50     if(x == t) {
 51         //printf("x == t maxF = %d 
", maxF);
 52         return maxF;
 53     }
 54     int ans = 0;
 55     for(int i = cur[x]; i; i = edge[i].nex) {
 56         int y = edge[i].v;
 57         if(dis[y] != dis[x] + 1 || !edge[i].c) continue;
 58         int temp = DFS(y, t, std::min(maxF - ans, edge[i].c));
 59         //printf("DFS %d -> %d  min %d %d  
", x, y, maxF - ans, edge[i].c);
 60         if(!temp) dis[y] = INF;
 61         ans += temp;
 62         edge[i].c -= temp;
 63         edge[i ^ 1].c += temp;
 64         if(ans == maxF) break;
 65         cur[x] = i;
 66     }
 67     return ans;
 68 }
 69 
 70 inline int solve(int s, int t) {
 71     int ans = 0;
 72     while(BFS(s, t)) {
 73         memcpy(cur + 1, e + 1, cnt * sizeof(int));
 74         //int temp = DFS(s, t, INF);
 75         //printf("ans += %d 
", temp);
 76         ans += DFS(s, t, INF);
 77 
 78     }
 79     return ans;
 80 }
 81 
 82 inline bool valid(int x, int y) {
 83     if(!x || !y || x > n || y > n) return false;
 84     return y >= x;
 85 }
 86 
 87 int main() {
 88     scanf("%d%d", &n, &m);
 89     int s = 1, t = 2, ans = 0;
 90     cnt = t;
 91     for(int i = 1; i <= n; i++) {
 92         scanf("%d", &a[i]);
 93     }
 94     for(int i = 1; i <= n; i++) {
 95         for(int j = i; j <= n; j++) {
 96             scanf("%d", &d[i][j]);
 97             id[i][j] = ++cnt;
 98             if(d[i][j] > 0) {
 99                 add(s, id[i][j], d[i][j]);
100                 ans += d[i][j];
101             }
102             else if(d[i][j] < 0) {
103                 add(id[i][j], t, -d[i][j]);
104             }
105             if(valid(i, j - 1)) {
106                 add(id[i][j], id[i][j - 1], INF);
107             }
108             if(valid(i - 1, j)) {
109                 add(id[i - 1][j], id[i][j], INF);
110             }
111         }
112         add(id[i][i], ++cnt, INF);
113         add(cnt, t, a[i]);
114         if(m) {
115             if(!use[a[i]]) {
116                 use[a[i]] = ++cnt;
117                 add(cnt - 1, cnt, INF);
118                 add(cnt, t, a[i] * a[i]);
119             }
120             else {
121                 add(cnt, use[a[i]], INF);
122             }
123         }
124     }
125 
126     printf("%d", ans - solve(s, t));
127     return 0;
128 }

AC代码

去膜了一下官方题解，发现这SB题没有状压做法......当m = 0的时候可以直接区间DP预处理每一段全选的最大价值，然后再在序列上按照断点DP。

如果能够重复计算一个子串的贡献呢？不会......

BZOJ3996 线性代数

题意：给出一个N * N的矩阵B和一个1 * N的矩阵C。求出一个1 * N的01矩阵A，使得D = (A * B - C) * A^T最大。输出D。

解：暴力推一波式子，发现D = ∑(∑(A_iA_jB_ij) - A_iC_i)，由于A是01矩阵，所以相当于选出若干个B，如果B_ij被选，那么-C_i和-C_j也会被选。求最大价值。最大权闭合子图。

  1 #include <bits/stdc++.h>
  2 
  3 const int N = 260010, INF = 0x3f3f3f3f;
  4 
  5 struct Edge {
  6     int nex, v, c;
  7     Edge(int NEX = 0, int V = 0, int C = 0) {
  8         nex = NEX;
  9         v = V;
 10         c = C;
 11     }
 12 }edge[2000010]; int tp = 1;
 13 
 14 int e[N], cur[N], d[N], vis[N], Time, cnt, n;
 15 std::queue<int> Q;
 16 
 17 inline void add(int x, int y, int z) {
 18     //printf("add %d %d 
", x, y);
 19     edge[++tp] = Edge(e[x], y, z);
 20     e[x] = tp;
 21     edge[++tp] = Edge(e[y], x, 0);
 22     e[y] = tp;
 23     return;
 24 }
 25 
 26 inline bool BFS(int s, int t) {
 27     Time++;
 28     vis[s] = Time;
 29     d[s] = 0;
 30     Q.push(s);
 31     while(Q.size()) {
 32         int x = Q.front();
 33         Q.pop();
 34         for(int i = e[x]; i; i = edge[i].nex) {
 35             int y = edge[i].v;
 36             if(vis[y] == Time || !edge[i].c) {
 37                 continue;
 38             }
 39             vis[y] = Time;
 40             d[y] = d[x] + 1;
 41             Q.push(y);
 42         }
 43     }
 44     return vis[t] == Time;
 45 }
 46 
 47 int DFS(int x, int t, int maxF) {
 48     if(x == t) {
 49         return maxF;
 50     }
 51     int ans = 0;
 52     for(int i = cur[x]; i; i = edge[i].nex) {
 53         int y = edge[i].v;
 54         if(d[y] != d[x] + 1 || !edge[i].c) continue;
 55         int temp = DFS(y, t, std::min(maxF - ans, edge[i].c));
 56         if(!temp) d[y] = INF;
 57         ans += temp;
 58         edge[i].c -= temp;
 59         edge[i ^ 1].c += temp;
 60         if(ans == maxF) break;
 61         cur[x] = i;
 62     }
 63     return ans;
 64 }
 65 
 66 inline int solve(int s, int t) {
 67     int ans = 0;
 68     while(BFS(s, t)) {
 69         memcpy(cur + 1, e + 1, cnt * sizeof(int));
 70         ans += DFS(s, t, INF);
 71     }
 72     return ans;
 73 }
 74 
 75 inline int id(int i, int j) {
 76     return (i - 1) * n + j;
 77 }
 78 
 79 int main() {
 80     scanf("%d", &n);
 81     int s = n * (n + 1) + 1, t = s + 1, ans = 0, x;
 82     cnt = t;
 83     for(int i = 1; i <= n; i++) {
 84         for(int j = 1; j <= n; j++) {
 85             scanf("%d", &x);
 86             if(x > 0) {
 87                 add(s, id(i, j), x);
 88                 ans += x;
 89             }
 90             else if(x < 0) {
 91                 add(id(i, j), t, -x);
 92             }
 93             add(id(i, j), id(n + 1, i), INF);
 94             add(id(i, j), id(n + 1, j), INF);
 95         }
 96     }
 97     for(int i = 1; i <= n; i++) {
 98         scanf("%d", &x);
 99         x = -x;
100         if(x > 0) {
101             add(s, id(n + 1, i), x);
102             ans += x;
103         }
104         else if(x < 0) {
105             add(id(n + 1, i), t, -x);
106         }
107     }
108 
109     printf("%d", ans - solve(s, t));
110     return 0;
111 }

AC代码

BZOJ3218 a + b Problem

题意：给你个序列，你要黑白染色。分别有贡献。每个位置还有权值。如果某个黑色位置前面有白色位置，且那个白色位置的权值在一个特定的限制中，那么要减去一个代价。求最大贡献。

解：按照前面几题最小割的套路，我们要把每个点向它前面符合条件的白点连边。注意到这些白点是二维矩阵内的，于是主席树优化一下。

  1 #include <bits/stdc++.h>
  2 
  3 typedef long long LL;
  4 const int N = 200010;
  5 const LL INF = 0x3f3f3f3f3f3f3f3fll;
  6 
  7 struct Edge {
  8     int nex, v;
  9     LL c;
 10     Edge(int NEX = 0, int V = 0, LL C = 0) {
 11         nex = NEX;
 12         v = V;
 13         c = C;
 14     }
 15 }edge[500010]; int tp = 1;
 16 
 17 int e[N], cur[N], d[N], vis[N], Time, cnt;
 18 std::queue<int> Q;
 19 int n, a[N], lc[N], rc[N], rt[N], last[N];
 20 LL b[N], w[N], c[N];
 21 int ls[N], rs[N], X[N], xx;
 22 
 23 inline void add(int x, int y, LL z) {
 24     //if(x == n * 2 + 1 || y == n * 2 + 2) {
 25         //printf("%d %d %lld 
", x, y, z);
 26     //}
 27     edge[++tp] = Edge(e[x], y, z);
 28     e[x] = tp;
 29     edge[++tp] = Edge(e[y], x, 0);
 30     e[y] = tp;
 31     return;
 32 }
 33 
 34 inline bool BFS(int s, int t) {
 35     Time++;
 36     vis[s] = Time;
 37     d[s] = 0;
 38     Q.push(s);
 39     while(Q.size()) {
 40         int x = Q.front();
 41         Q.pop();
 42         for(int i = e[x]; i; i = edge[i].nex) {
 43             int y = edge[i].v;
 44             if(vis[y] == Time || !edge[i].c) {
 45                 continue;
 46             }
 47             vis[y] = Time;
 48             d[y] = d[x] + 1;
 49             Q.push(y);
 50         }
 51     }
 52     return vis[t] == Time;
 53 }
 54 
 55 LL DFS(int x, int t, LL maxF) {
 56     if(x == t) {
 57         return maxF;
 58     }
 59     LL ans = 0;
 60     for(int i = cur[x]; i; i = edge[i].nex) {
 61         int y = edge[i].v;
 62         if(d[y] != d[x] + 1 || !edge[i].c) continue;
 63         LL temp = DFS(y, t, std::min(maxF - ans, edge[i].c));
 64         if(!temp) d[y] = INF;
 65         ans += temp;
 66         edge[i].c -= temp;
 67         edge[i ^ 1].c += temp;
 68         if(ans == maxF) break;
 69         cur[x] = i;
 70     }
 71     return ans;
 72 }
 73 
 74 inline LL solve(int s, int t) {
 75     LL ans = 0;
 76     while(BFS(s, t)) {
 77         memcpy(cur + 1, e + 1, cnt * sizeof(int));
 78         ans += DFS(s, t, INF);
 79     }
 80     return ans;
 81 }
 82 
 83 void insert(int &x, int y, int p, int v, int l, int r) {
 84     if(!x || x == y) {
 85         x = ++cnt;
 86         ls[x] = ls[y];
 87         rs[x] = rs[y];
 88     }
 89     //printf("insert x = %d 
", x);
 90     if(l == r) {
 91         if(last[r]) {
 92             add(x, last[r], INF);
 93         }
 94         add(x, v, INF);
 95         last[r] = x;
 96         return;
 97     }
 98     int mid = (l + r) >> 1;
 99     if(p <= mid) {
100         insert(ls[x], ls[y], p, v, l, mid);
101     }
102     else {
103         insert(rs[x], rs[y], p, v, mid + 1, r);
104     }
105     return;
106 }
107 
108 void link(int L, int R, int v, int l, int r, int o) {
109     if(!o) return;
110     if(L <= l && r <= R) {
111         add(v, o, INF);
112         return;
113     }
114     int mid = (l + r) >> 1;
115     if(L <= mid) link(L, R, v, l, mid, ls[o]);
116     if(mid < R) link(L, R, v, mid + 1, r, rs[o]);
117     return;
118 }
119 
120 void build(int l, int r, int o) {
121     if(!o || vis[o]) return;
122     vis[o] = 1;
123     int mid = (l + r) >> 1;
124     if(ls[o]) {
125         add(o, ls[o], INF);
126         build(l, mid, ls[o]);
127     }
128     if(rs[o]) {
129         add(o, rs[o], INF);
130         build(mid + 1, r, rs[o]);
131     }
132     return;
133 }
134 
135 int main() {
136 
137     //printf("%d 
", (sizeof(edge) + sizeof(e) * 12 + sizeof(b) * 3) / 1048576);
138 
139     scanf("%d", &n);
140     int s = n * 2 + 1, t = s + 1;
141     cnt = t;
142     for(int i = 1; i <= n; i++) {
143         scanf("%d%lld%lld%d%d%lld", &a[i], &b[i], &w[i], &lc[i], &rc[i], &c[i]);
144         X[i] = a[i];
145     }
146     std::sort(X + 1, X + n + 1);
147     xx = std::unique(X + 1, X + n + 1) - X - 1;
148     for(int i = 1; i <= n; i++) {
149         a[i] = std::lower_bound(X + 1, X + xx + 1, a[i]) - X;
150         lc[i] = std::lower_bound(X + 1, X + xx + 1, lc[i]) - X;
151         rc[i] = std::upper_bound(X + 1, X + xx + 1, rc[i]) - X - 1;
152         if(i < n) {
153             insert(rt[i], rt[i - 1], a[i], i, 1, xx);
154             //printf("i = %d 
", i);
155         }
156     }
157 
158     LL ans = 0;
159 
160     for(int i = 1; i <= n; i++) {
161         add(s, i, b[i]);
162         add(i, t, w[i]);
163         ans += b[i] + w[i];
164         if(lc[i] > rc[i]) {
165             continue;
166         }
167         if(i > 1) {
168             add(i, i + n, c[i]);
169             link(lc[i], rc[i], i + n, 1, xx, rt[i - 1]);
170         }
171     }
172     ++Time;
173     for(int i = 1; i < n; i++) build(1, xx, rt[i]);
174 
175     printf("%lld
", ans - solve(s, t));
176     return 0;
177 }

AC代码

BZOJ4501 旅行

题意：给定一张有向图，你要删去一些边，使得从1出发随机前进的期望长度最长。有些限制是删a边则必删b边，保证a和b起点相同。

解：发现这个起点相同很妙啊，那么这些点就是互相独立的了。然后逆拓扑序考虑。对每个点来说，我们要让∑(a_iv_i) / ∑a_i最大。

发现这是01分数规划问题，于是二分答案，就是求∑a_i(v_i - k)的最大值，且有些限制是不选a则一定不选b。于是先全部选了，就是选a一定选b，求最小权闭合子图即可。

  1 #include <bits/stdc++.h>
  2 
  3 typedef long long LL;
  4 const int N = 20010;
  5 const double INF = 1e9;
  6 const double eps = 1e-12;
  7 
  8 struct Edge {
  9     int nex, v;
 10     double c;
 11     Edge(int NEX = 0, int V = 0, double C = 0) {
 12         nex = NEX;
 13         v = V;
 14         c = C;
 15     }
 16 }edge[500010]; int tp = 1;
 17 
 18 struct Node {
 19     int l, r;
 20 }node[N];
 21 
 22 int e[N], cur[N], d[N], vis[N], Time, cnt;
 23 std::queue<int> Q;
 24 std::vector<int> v[N], v2[N];
 25 double val[N], Val[N];
 26 int n, m, K, e2[N], in[N], topo[N], tp2, posu[N], posv[N], in2[N], id[N];
 27 Edge edge2[N];
 28 
 29 inline void add(int x, int y, double z) {
 30     edge[++tp] = Edge(e[x], y, z);
 31     e[x] = tp;
 32     edge[++tp] = Edge(e[y], x, 0);
 33     e[y] = tp;
 34     return;
 35 }
 36 
 37 inline bool BFS(int s, int t) {
 38     Time++;
 39     vis[s] = Time;
 40     d[s] = 0;
 41     Q.push(s);
 42     while(Q.size()) {
 43         int x = Q.front();
 44         Q.pop();
 45         for(int i = e[x]; i; i = edge[i].nex) {
 46             int y = edge[i].v;
 47             if(vis[y] == Time || edge[i].c < eps) {
 48                 continue;
 49             }
 50             vis[y] = Time;
 51             d[y] = d[x] + 1;
 52             Q.push(y);
 53         }
 54     }
 55     return vis[t] == Time;
 56 }
 57 
 58 double DFS(int x, int t, double maxF) {
 59     if(x == t) {
 60         return maxF;
 61     }
 62     double ans = 0;
 63     for(int i = cur[x]; i; i = edge[i].nex) {
 64         int y = edge[i].v;
 65         if(d[y] != d[x] + 1 || edge[i].c < eps) continue;
 66         double temp = DFS(y, t, std::min(maxF - ans, edge[i].c));
 67         if(temp < eps) d[y] = INF;
 68         ans += temp;
 69         edge[i].c -= temp;
 70         edge[i ^ 1].c += temp;
 71         if(fabs(ans - maxF) < eps) break;
 72         cur[x] = i;
 73     }
 74     return ans;
 75 }
 76 
 77 inline double solve(int s, int t) {
 78     double ans = 0;
 79     while(BFS(s, t)) {
 80         memcpy(cur + 1, e + 1, cnt * sizeof(int));
 81         ans += DFS(s, t, INF);
 82     }
 83     return ans;
 84 }
 85 
 86 /// --------------------------- Dinic -------------------------------------
 87 
 88 inline void add2(int x, int y) {
 89     edge2[++tp2] = Edge(e2[x], y);
 90     e2[x] = tp2;
 91     return;
 92 }
 93 
 94 inline bool check(int x, double k) {
 95 
 96     tp = 1;
 97     memset(e + 1, 0, cnt * sizeof(int));
 98 
 99     int s = 1, t = 2;
100     cnt = t;
101     double ans = 0, tans = 0;
102     for(int i = 0; i < v2[x].size(); i++) {
103         int y = posv[v2[x][i]];
104         Val[y] = val[y] - k;
105         tans += Val[y];
106         Val[y] = -Val[y];
107         id[v2[x][i]] = ++cnt;
108         if(Val[y] > eps) {
109             add(s, cnt, Val[y]);
110             ans += Val[y];
111         }
112         else if(Val[y] < -eps) {
113             add(cnt, t, -Val[y]);
114         }
115     }
116 
117     for(int i = 0; i < v[x].size(); i++) {
118         int t = v[x][i];
119         add(id[node[t].l], id[node[t].r], INF);
120     }
121 
122     ans -= solve(s, t);
123     tans += ans;
124     return tans > eps;
125 }
126 /*
127 3 3 1
128 1 2
129 1 3
130 2 3
131 1 2
132 
133 */
134 int main() {
135 
136     scanf("%d%d%d", &n, &m, &K);
137     for(int i = 1, x, y; i <= m; i++) {
138         scanf("%d%d", &x, &y);
139         add2(x, y);
140         v2[x].push_back(i);
141         posu[i] = x;
142         posv[i] = y;
143         in[y]++;
144         in2[x]++;
145     }
146     for(int i = 1, x, y; i <= K; i++) {
147         scanf("%d%d", &node[i].r, &node[i].l);
148         v[posu[node[i].l]].push_back(i);
149     }
150 
151     for(int i = 1; i <= n; i++) {
152         if(!in[i]) Q.push(i);
153     }
154     while(Q.size()) {
155         int x = Q.front();
156         Q.pop();
157         topo[++cnt] = x;
158         for(int i = e2[x]; i; i = edge2[i].nex) {
159             int y = edge2[i].v;
160             in[y]--;
161             if(!in[y]) {
162                 Q.push(y);
163             }
164         }
165     }
166 
167     cnt = 0;
168     for(int alpha = n; alpha >= 1; alpha--) {
169         int x = topo[alpha];
170         if(!in2[x]) {
171             continue;
172         }
173         if(!v[x].size()) {
174             for(int i = e2[x]; i; i = edge2[i].nex) {
175                 int y = edge2[i].v;
176                 val[x] = std::max(val[x], val[y]);
177             }
178             val[x] = val[x] + 1;
179             continue;
180         }
181         double l = 0, r = m;
182         for(int beta = 1; beta <= 40; beta++) {
183             double mid = (l + r) / 2;
184             if(check(x, mid)) {
185                 l = mid;
186             }
187             else {
188                 r = mid;
189             }
190         }
191         val[x] = r + 1;
192         if(x == 1) break;
193     }
194     printf("%.6f
", val[1]);
195     return 0;
196 }

AC代码

查看全文

相关阅读:
aws-rds for mysql 5.7.34搭建备库
 Redis 未授权访问漏洞利用总结(转)
mongoexport/mongimport命令详解
 mongodump/mongorestore命令详解
 redis stream类型常用命令
 system_time_zone参数值由来
 MySQL加密解密函数AES_ENCRYPT AES_DECRYPT
MySQL开启SSL加密
 MDL锁获取顺序和优先先
 explicit_defaults_for_timestamp 参数说明

原文地址：https://www.cnblogs.com/huyufeifei/p/10729965.html