【BZOJ1834】 [ZJOI2010]network 网络扩容

给定一张有向图，每条边都有一个容量C和一个扩容费用W。这里扩容费用是指将容量扩大1所需的费用。求： 1、在不扩容的情况下，1到N的最大流； 2、将1到N的最大流增加K所需的最小扩容费用。

输入文件的第一行包含三个整数N,M,K，表示有向图的点数、边数以及所需要增加的流量。接下来的M行每行包含四个整数u,v,C,W，表示一条从u到v，容量为C，扩容费用为W的边。

输出文件一行包含两个整数，分别表示问题1和问题2的答案。

5 8 2
1 2 5 8
2 5 9 9
5 1 6 2
5 1 1 8
1 2 8 7
2 5 4 9
1 2 1 1
1 4 2 1

13 19
30%的数据中，N<=100
100%的数据中，N<=1000，M<=5000，K<=10

Solution

第一问可以直接用最大流跑。

第二问新建源然后向旧源连一条容量为K，费用0的边。其他就是原来的连容量c，费用0，新加的连容量inf，费用W。最后跑费用流即可。

Code

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <algorithm>
  4 #include <cmath>
  5 #include <queue>
  6 
  7 #ifdef WIN32
  8     #define LL "%I64d"
  9 #else
 10     #define LL "%lld"
 11 #endif
 12 
 13 #ifdef CT
 14     #define debug(...) printf(__VA_ARGS__)
 15     #define setfile() 
 16 #else
 17     #define debug(...)
 18     #define filename ""
 19     #define setfile() freopen(filename".in", "r", stdin); freopen(filename".out", "w", stdout)
 20 #endif
 21 
 22 #define R register
 23 #define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T) ? EOF : *S++)
 24 #define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b))
 25 #define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b))
 26 #define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0)
 27 #define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0)
 28 #define cabs(_x) ((_x) < 0 ? (- (_x)) : (_x))
 29 char B[1 << 15], *S = B, *T = B;
 30 inline int F()
 31 {
 32     R char ch; R int cnt = 0; R bool minus = 0;
 33     while (ch = getc(), (ch < '0' || ch > '9') && ch != '-') ;
 34     ch == '-' ? minus = 1 : cnt = ch - '0';
 35     while (ch = getc(), ch >= '0' && ch <= '9') cnt = cnt * 10 + ch - '0';
 36     return minus ? -cnt : cnt;
 37 }
 38 #define maxn 1010
 39 #define maxm 100010
 40 struct edge
 41 {
 42     int a, b, w, c;
 43 }ee[maxm];
 44 struct Edge
 45 {
 46     Edge *next, *rev;
 47     int from, to, cap, cost;
 48 }*last[maxn], e[maxm], *ecnt = e, *cur[maxn], *prev[maxn];
 49 inline void link(R int a, R int b, R int w, R int c)
 50 {
 51     *++ecnt = (Edge) {last[a], ecnt + 1, a, b, w, c}; last[a] = ecnt;
 52     *++ecnt = (Edge) {last[b], ecnt - 1, b, a, 0, -c}; last[b] = ecnt;
 53 }
 54 #define inf 0x7fffffff
 55 int s, t, dep[maxn], ans, d[maxn];
 56 std::queue<int> q;
 57 inline bool bfs()
 58 {
 59     memset(dep, -1, sizeof (dep));
 60     dep[t] = 0; q.push(t);
 61     while (!q.empty())
 62     {
 63         R int now = q.front(); q.pop();
 64         for (R Edge *iter = last[now]; iter; iter = iter -> next)
 65         {
 66             R int pre = iter -> to;
 67             if (iter -> rev -> cap && dep[pre] == -1)
 68             {
 69                 dep[pre] = dep[now] + 1;
 70                 q.push(pre);
 71             }
 72         }
 73     }
 74     return dep[s] != -1;
 75 }
 76 int dfs(R int x, R int f)
 77 {
 78     if (x == t) return f;
 79     R int used = 0;
 80     for (R Edge* &iter = cur[x]; iter; iter = iter -> next)
 81     {
 82         R int pre = iter -> to;
 83         if (iter -> cap && dep[x] == dep[pre] + 1)
 84         {
 85             R int v = dfs(pre, dmin(iter -> cap, f - used));
 86             iter -> cap -= v;
 87             iter -> rev -> cap += v;
 88             used += v;
 89             if (f == used) return f;
 90         }
 91     }
 92     if (!used) dep[x] = -1;
 93     return used;
 94 }
 95 inline void dinic()
 96 {
 97     while (bfs())
 98     {
 99         memcpy(cur, last, sizeof last);
100         ans += dfs(s, inf);
101     }
102 }
103 bool inq[maxn];
104 inline bool spfa()
105 {
106     for (R int i = 1; i <= t; ++i) d[i] = inf;
107     q.push(s);
108     while (!q.empty())
109     {
110         R int now = q.front(); q.pop(); inq[now] = 0;
111         for (R Edge *iter = last[now]; iter; iter = iter -> next)
112         {
113             R int pre = iter -> to;
114             if (iter -> cap && d[pre] > d[now] + iter -> cost)
115             {
116                 d[pre] = d[now] + iter -> cost;
117                 prev[pre] = iter;
118                 if (!inq[pre])
119                 {
120                     q.push(pre);
121                     inq[pre] = 1;
122                 }
123             }
124         }
125     }
126     return d[t] != inf;
127 }
128 inline void mcmf()
129 {
130     R int x = inf;
131     for (R Edge *iter = prev[t]; iter; iter = prev[iter -> from])
132         cmin(x, iter -> cap);
133     for (R Edge *iter = prev[t]; iter; iter = prev[iter -> from])
134     {
135         iter -> cap -= x;
136         iter -> rev -> cap += x;
137         ans += x * iter -> cost;
138     }
139 }
140 int main()
141 {
142 //    setfile();
143     R int n = F(), m = F(), k = F();
144     for (R int i = 1; i <= m; ++i)
145     {
146         R int a = F(), b = F(), w = F(), c = F();
147         link(a, b, w, 0); ee[i] = (edge) {a, b, w, c};
148     }
149     s = 1; t = n;
150     dinic();
151     printf("%d ", ans );
152     ans = 0;
153     for (R int i = 1; i <= m; ++i) link(ee[i].a, ee[i].b, inf, ee[i].c);
154     s = 0; t = n;
155     link(s, 1, k, 0);
156     while (spfa()) mcmf();
157     printf("%d
", ans );
158     return 0;
159 }

【BZOJ1834】 [ZJOI2010]network 网络扩容

Description

Input

Output

Sample Input

Sample Output

Solution

Code