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＂新生赛＂长沙理工大学程序设计竞赛

B、happy card

列出所有条件：人数

其实算法还可以再进行时间的优化，将其化为

n^2 算法。

实际上人数是一个无用条件，我们知道总的快乐牌数，那么人数一定会大于等于这个数。这种情况的所有必要条件就是：总的快乐牌

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <string>
 4 #include <string.h>
 5 #include <vector>
 6 #include <map>
 7 #include <stack>
 8 #include <set>
 9 #include <queue>
10 #include <math.h>
11 #include <cstdio>
12 #include <time.h>
13  
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18  
19 const int maxn = 1e6 + 10;
20 const LL MOD = 1e9 + 7;
21  
22 LL cnt;
23 int n,k,s;
24 LL dp[1010];
25  
26  
27  
28 int main() {
29     scanf("%d%d%d",&n,&k,&s);
30     for (int i = 1;i <= n;i++) {
31         int val;
32         scanf("%d",&val);
33         if (val == s)
34             cnt++;
35     }
36     for (int i = 1;i <= k;i++) {
37         int val;
38         scanf("%d", &val);
39         for (int j = 0;j <= cnt;j++) {
40             if (j >= i)
41                 dp[j] = std::max(dp[j],dp[j-i]+val);
42         }
43     }
44     printf("%lld
",dp[cnt]);
45     return 0;
46 }

Ackerman

D、巫妖王的远征

典型的 BFS 的题目，主要问题是如何处理 传送门 和 休息时间 的问题。

传送门：就是如果到了可以传送的地方我们就把它传送后的地点直接入队就好了

休息时间：最后找到了直接统一处理就好了

  1 #include <iostream>
  2 #include <algorithm>
  3 #include <string>
  4 #include <string.h>
  5 #include <vector>
  6 #include <map>
  7 #include <stack>
  8 #include <set>
  9 #include <queue>
 10 #include <math.h>
 11 #include <cstdio>
 12 #include <time.h>
 13 
 14 #define LL long long
 15 #define INF 0x3f3f3f3f
 16 #define ls nod<<1
 17 #define rs (nod<<1)+1
 18 
 19 const int maxn = 3e5 + 10;
 20 const LL MOD = 1e9 + 7;
 21 
 22 char map[1010][1010];
 23 int vis[1010][1010];
 24 int trans[1010][1010];
 25 int dx[] = {0,0,1,-1};
 26 int dy[] = {1,-1,0,0};
 27 int sx,sy,ex,ey;
 28 
 29 struct Node {
 30     int x,y,step;
 31 }a,b;
 32 
 33 std::queue<Node> q;
 34 int main() {
 35     int n,m;
 36     scanf("%d%d",&n,&m);
 37     for (int i = 1;i <= n;i++)
 38         scanf("%s",map[i]+1);
 39     for (int i = 1;i <= n;i++) {
 40         for (int j = 1;j <= m;j++) {
 41             if (map[i][j] == 'S') {
 42                 a.x = i;
 43                 a.y = j;
 44                 a.step = 0;
 45                 q.push(a);
 46             }
 47             if (map[i][j] == 'T') {
 48                 ex = i;
 49                 ey = j;
 50             }
 51         }
 52     }
 53     int k;
 54     scanf("%d",&k);
 55     for (int i = 1;i <= k;i++) {
 56         scanf("%d%d%d%d",&trans[i][0],&trans[i][1],&trans[i][2],&trans[i][3]);
 57         if (trans[i][0] <= 0 || trans[i][0] > n || trans[i][1] <= 0 || trans[i][1] > m || trans[i][2] <= 0 || trans[i][2] > n || trans[i][3] <= 0 || trans[i][3] > m)
 58         {
 59             trans[i][0] = -1;
 60             trans[i][2] = -1;
 61         }
 62         else if (map[trans[i][0]][trans[i][1]] == '#' || map[trans[i][2]][trans[i][3]] == '#')
 63         {
 64             trans[i][0] = -1;
 65             trans[i][2] = -1;
 66         }
 67     }
 68     for (int i = 1;i <= k;i++) {
 69         if ((a.x == trans[i][0] && a.y == trans[i][1]) && (ex == trans[i][2] && ey == trans[i][3]))
 70         {
 71             printf("0
");
 72             return 0;
 73         }
 74         else if ((a.x == trans[i][2] && a.y == trans[i][3]) && (ex == trans[i][0] && ey == trans[i][1]))
 75         {
 76             printf("0
");
 77             return 0;
 78         }
 79     }
 80     bool flag = false;
 81     int ans = 0;
 82     while (!q.empty()) {
 83         a = q.front();
 84         q.pop();
 85         if (a.x == ex && a.y == ey) {
 86             flag = true;
 87             ans = a.step + a.step / 8;
 88             if (a.step % 8 == 0)
 89                 ans--;
 90             printf("%d
",ans);
 91             break;
 92         }
 93         for (int i = 0;i < 4;i++) {
 94             int x = a.x + dx[i];
 95             int y = a.y + dy[i];
 96             if (x <= 0 || y <= 0 || x > n || y > m || vis[x][y] || map[x][y] == '#')
 97                 continue;
 98             for (int j = 1;j <= k;j++) {
 99                 if (x == trans[j][0] && y == trans[j][1])
100                 {
101                     b.x = trans[j][2];
102                     b.y = trans[j][3];
103                     b.step = a.step + 1;
104                     q.push(b);
105                     vis[b.x][b.y] = 1;
106                 }
107                 else if (x == trans[j][2] && y == trans[j][3])
108                 {
109                     b.x = trans[j][0];
110                     b.y = trans[j][1];
111                     b.step = a.step + 1;
112                     q.push(b);
113                     vis[b.x][b.y] = 1;
114                 }
115             }
116             vis[x][y] = 1;
117             b.x = x;
118             b.y = y;
119             b.step = a.step + 1;
120             q.push(b);
121         }
122     }
123     if (!flag)
124         printf("-1
");
125     return 0;
126 }

Ackerman

I、洛克王国之域外大逃杀

http://acm.csust.edu.cn/problem/3021
我们可以先只考虑暴力，也就是先20遍bfs遍历所有的魔物，然后在地图上记录一下每个魔物抵达的时间和该时间点的最大攻击。

接下来由于最多只有两个宠物，我们跑4次bfs就好了，洛克到1号宠物再到2号，洛克到2号再到1号。于是这题就愉快的暴力掉了4000ms+。然后就是考虑优化了。

其实也不需要优化太多，我们可以将所有魔物全部放进队列中一起跑，然后稍微剪个枝就可以过了。我们知道如果一个魔物后抵达A，而它的攻击又弱于或者等于前一个抵达的魔物，那么我们就可以将这个状态给舍弃掉了。1500ms

总结下这道题目：

主要的难点我觉得还是在处理魔物的移动，某个位置可以让两个不同的魔物跑，怎么去避免去重的时候出错。

然后就是第二个 bfs 的时候，处理洛克的时候，官方给的 code 处理的也很巧妙

然后能用 bool 类型就用 bool ，不要乱开 int 会爆内存

  1 #include <iostream>
  2 #include <algorithm>
  3 #include <string>
  4 #include <string.h>
  5 #include <vector>
  6 #include <map>
  7 #include <stack>
  8 #include <set>
  9 #include <queue>
 10 #include <math.h>
 11 #include <cstdio>
 12 #include <time.h>
 13 
 14 #define LL long long
 15 #define INF 0x3f3f3f3f
 16 #define ls nod<<1
 17 #define rs (nod<<1)+1
 18 
 19 const int maxn = 1e5 + 10;
 20 const LL MOD = 1e9 + 7;
 21 
 22 int n,m,stak;
 23 int cnte,cnta;
 24 char mp[1010][1010];
 25 int dx[] = {0,0,1,-1};
 26 int dy[] = {1,-1,0,0};
 27 bool vis[1010][1010][25];
 28 bool visa[1010][1010];
 29 
 30 struct node1 {
 31     int x,y,atk,p;
 32 }evil[25],ang[20];
 33 
 34 struct node2 {
 35     int x,y,atk,p,t;
 36 };
 37 
 38 struct node3 {
 39     int t,atk;
 40 }ev[1010][1010][25];
 41 
 42 bool ok(int x,int y) {
 43     if (x>=1 && x<=n && y>=1 && y<=m)
 44         return true;
 45     return false;
 46 }
 47 
 48 void bfs() {
 49     std::queue<node2> q;
 50     int t = 0;
 51     for (int i = 1;i <= cnte;i++) {
 52         node2 now;
 53         now.x = evil[i].x;
 54         now.y = evil[i].y;
 55         now.p = evil[i].p;
 56         now.atk = evil[i].atk;
 57         now.t = t;
 58         q.push(now);
 59         ev[now.x][now.y][i].t = 0;
 60         ev[now.x][now.y][i].atk = now.atk;
 61     }
 62     while (!q.empty()) {
 63         node2 now = q.front();
 64         q.pop();
 65         int x = now.x,y = now.y,p = now.p;
 66         if (vis[x][y][p])
 67             continue;
 68         vis[x][y][p] = 1;
 69         for (int i = 0;i < 4;i++) {
 70             int xx = x + dx[i];
 71             int yy = y + dy[i];
 72             t = now.t + 2;
 73             if (!ok(xx,yy))
 74                 continue;
 75             if (vis[xx][yy][p])
 76                 continue;
 77             int j;
 78             for (j = 1;j <= cnte;j++) {
 79                 if (ev[xx][yy][j].atk >= now.atk && ev[xx][yy][j].t <= t)
 80                     break;
 81             }
 82             if (j <= cnte)
 83                 continue;
 84             node2 now2 = now;
 85             now2.t = t;
 86             now2.x = xx;
 87             now2.y = yy;
 88             ev[xx][yy][p].atk = now.atk;
 89             ev[xx][yy][p].t = t;
 90             q.push(now2);
 91         }
 92     }
 93 }
 94 
 95 node3 bfs2(int sx,int sy,int ex,int ey,int id,int stk,int base) {
 96     if (sx == ex && sy == ey)
 97         return {0,stk};
 98     std::queue<node2> q;
 99     node2 now;
100     now.x = sx;
101     now.y = sy;
102     now.atk = stk;
103     now.t = base;
104     q.push(now);
105     memset(visa,0, sizeof(visa));
106     while (!q.empty()) {
107         now = q.front();
108         q.pop();
109         int x = now.x,y = now.y;
110         if (visa[x][y])
111             continue;
112         visa[x][y] = 1;
113         for (int i = 0;i < 4;i++) {
114             int xx = x + dx[i];
115             int yy = y + dy[i];
116             if (visa[xx][yy])
117                 continue;
118             if (!ok(xx,yy))
119                 continue;
120             int j;
121             for (j = 1;j <= cnte;j++) {
122                 if (ev[xx][yy][j].t <= now.t+1 && now.atk <= ev[xx][yy][j].atk)
123                     break;
124             }
125             if (j <= cnte)
126                 continue;
127             if (xx == ex && yy == ey)
128                 return {now.t+1,now.atk+ang[id].atk};
129             node2 now2 = now;
130             now2.x = xx;
131             now2.y = yy;
132             now2.t += 1;
133             q.push(now2);
134         }
135     }
136     return {INF,0};
137 }
138 
139 
140 int main() {
141     std::cin >> n >> m >> stak;
142     int sx,sy;
143     cnte = 1,cnta = 1;
144     for (int i = 1;i <= n;i++) {
145         for (int j = 1;j <= m;j++) {
146             std::cin >> mp[i][j];
147             if (mp[i][j] == '*') {
148                 evil[cnte].x = i;
149                 evil[cnte].y = j;
150                 evil[cnte].p = cnte;
151                 evil[cnte++].atk = 0;
152             }
153             if (mp[i][j] == '#') {
154                 ang[cnta].x = i;
155                 ang[cnta].y = j;
156                 ang[cnta].p = cnta;
157                 ang[cnta++].atk = 0;
158             }
159             if (mp[i][j] == 'L') {
160                 sx = i;
161                 sy = j;
162             }
163         }
164     }
165     cnte--;
166     cnta--;
167     for (int i = 1;i <= cnte;i++) {
168         int val;
169         std::cin >> val;
170         evil[i].atk = val;
171     }
172     for (int i = 1;i <= cnta;i++) {
173         int val;
174         std::cin >> val;
175         ang[i].atk = val;
176     }
177     if (!cnta) {
178         printf("0
");
179         return 0;
180     }
181     bfs();
182     node3 s1,s2,s3,s4;
183     s1.atk = s2.atk = s3.atk = s4.atk = 0;
184     s1.t = s2.t = s3.t = s4.t = INF;
185     if (cnta) {
186         if (cnta >= 1)
187             s1 = bfs2(sx,sy,ang[1].x,ang[1].y,1,stak,0);
188         if (cnta > 1) {
189             s2 = bfs2(ang[1].x,ang[1].y,ang[2].x,ang[2].y,2,s1.atk,s1.t);
190             s3 = bfs2(sx,sy,ang[2].x,ang[2].y,2,stak,0);
191             s4 = bfs2(ang[2].x,ang[2].y,ang[1].x,ang[1].y,1,s3.atk,s3.t);
192         }
193     }
194     int ans = INF;
195     if (cnta > 1)
196         ans = std::min(s2.t,s4.t);
197     else
198         ans = s1.t;
199     if (ans >= INF)
200         printf("you die!
");
201     else
202         printf("%d
",ans);
203     return 0;
204 }

Ackerman

J、jcc爱踢球

思维题。如果可以想到满二叉树的话那么这题应该就比较简单了。

由于每轮都是偶数只队伍，那么这就相当于一颗满二叉树了，我们经常看电视的时候每次比赛都基本会出现树状图。
我们每次只记录叶节点的编号（假设为

那么他们相遇也就是父节点相同的时候，而每找一次父节点计数器加一，则最后的计数器的值(记为 $2^S 为总的队伍数，则说明决赛相遇。$

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <string>
 4 #include <string.h>
 5 #include <vector>
 6 #include <map>
 7 #include <stack>
 8 #include <set>
 9 #include <queue>
10 #include <math.h>
11 #include <cstdio>
12 #include <time.h>
13 
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18 
19 const int maxn = 1e6 + 10;
20 const LL MOD = 1e9 + 7;
21 
22 int n,a,b;
23 
24 
25 int main() {
26     scanf("%d%d%d",&n,&a,&b);
27     if (a > b)
28         std::swap(a,b);
29     int s = 1;
30     while (a != b-1 || a % 2 == 0) {
31         a = (a + 1) / 2;
32         b = (b + 1) / 2;
33         s++;
34     }
35     if (n == pow(2,s))
36         printf("Final!
");
37     else
38         printf("%d
",s);
39     return 0;
40 }

Ackerman

L、这是最难的题

典型的直接对答案进行二分的题目，然后直接检查正确性就好了

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <string>
 4 #include <string.h>
 5 #include <vector>
 6 #include <map>
 7 #include <stack>
 8 #include <set>
 9 #include <queue>
10 #include <math.h>
11 #include <cstdio>
12 #include <time.h>
13 
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18 
19 const int maxn = 3e5 + 10;
20 const LL MOD = 1e9 + 7;
21 
22 
23 int n,m;
24 int a[maxn];
25 
26 bool check(int k) {
27     LL sum = 0;
28     for (int i = 1;i <= n;i++) {
29         sum += std::min(k,a[i]);
30     }
31     return sum >= (LL)k * m;
32 }
33 
34 int main() {
35     scanf("%d%d",&n,&m);
36     for (int i = 1;i <= n;i++)
37         scanf("%d",&a[i]);
38     std::sort(a+1,a+1+n);
39     int l = 0,r = 1e9;
40     int ans = 0;
41     while (l <= r) {
42         int mid = (l + r) >> 1;
43         if (check(mid)) {
44             ans = mid;
45             l = mid + 1;
46         }
47         else
48             r = mid - 1;
49     }
50     printf("%d
",ans);
51     return 0;
52 }

Ackerman

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原文地址：https://www.cnblogs.com/-Ackerman/p/12209097.html