[luoguP2763] 试题库问题（最大流）

传送门

 

每个类别和它所有的试题连一条权值为1的边。

增加一个超级源点s，s和每个类别连一条权值为选当前类别数量的边。

增加一个超级汇点t，每个试题和t连一条权值为1的边。

 

求最大流即可。

 

——代码

  1 #include <queue>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <iostream>
  5 #define min(x, y) ((x) < (y) ? (x) : (y))
  6 #define N 1100
  7 #define M 3000001
  8 
  9 int k, n, m, cnt, s, t, sum;
 10 int num[N], a[N][N];
 11 int head[N], to[M], next[M], val[M], dis[N], cur[N];
 12 
 13 inline int read()
 14 {
 15     int x = 0, f = 1;
 16     char ch = getchar();
 17     for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
 18     for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
 19     return x * f;
 20 }
 21 
 22 inline void add(int x, int y, int z)
 23 {
 24     to[cnt] = y;
 25     val[cnt] = z;
 26     next[cnt] = head[x];
 27     head[x] = cnt++;
 28 }
 29 
 30 inline bool bfs()
 31 {
 32     int i, u, v;
 33     std::queue <int> q;
 34     memset(dis, -1, sizeof(dis));
 35     q.push(s);
 36     dis[s] = 0;
 37     while(!q.empty())
 38     {
 39         u = q.front(), q.pop();
 40         for(i = head[u]; i ^ -1; i = next[i])
 41         {
 42             v = to[i];
 43             if(val[i] && dis[v] == -1)
 44             {
 45                 dis[v] = dis[u] + 1;
 46                 if(v == t) return 1;
 47                 q.push(v);
 48             }
 49         }
 50     }
 51     return 0;
 52 }
 53 
 54 inline int dfs(int u, int maxflow)
 55 {
 56     if(u == t) return maxflow;
 57     int i, v, d, ret = 0;
 58     for(i = cur[u]; i ^ -1; i = next[i])
 59     {
 60         v = to[i];
 61         if(val[i] && dis[v] == dis[u] + 1)
 62         {
 63             d = dfs(v, min(val[i], maxflow - ret));
 64             ret += d;
 65             cur[u] = i;
 66             val[i] -= d;
 67             val[i ^ 1] += d;
 68             if(ret == maxflow) return ret;
 69         }
 70     }
 71     return ret;
 72 }
 73 
 74 int main()
 75 {
 76     int i, j, p, x;
 77     k = read();
 78     n = read();
 79     s = 0, t = n + k + 1;
 80     for(i = 1; i <= k; i++) m += num[i] = read();
 81     for(i = 1; i <= n; i++)
 82     {
 83         p = read();
 84         for(j = 1; j <= p; j++) x = read(), a[x][++a[x][0]] = i;
 85     }
 86     memset(head, -1, sizeof(head));
 87     for(i = 1; i <= n; i++) add(i, t, 1), add(t, i, 0);
 88     for(i = 1; i <= k; i++) add(s, i + n, num[i]), add(i + n, s, 0);
 89     for(i = 1; i <= k; i++)
 90         for(j = 1; j <= a[i][0]; j++)
 91             add(i + n, a[i][j], 1), add(a[i][j], i + n, 0);
 92     while(bfs())
 93     {
 94         for(i = s; i <= t; i++) cur[i] = head[i];
 95         sum += dfs(s, 1e9);
 96     }
 97     if(sum ^ m)
 98     {
 99         puts("No Solution!");
100         return 0;
101     }
102     for(i = n + 1; i <= n + k; i++)
103     {
104         printf("%d:", i - n);
105         for(j = head[i]; j ^ -1; j = next[j])
106             if(!val[j] && to[j] ^ s)
107                 printf(" %d", to[j]);
108         puts("");
109     }
110     return 0;
111 }
View Code

 

posted @ 2017-06-02 16:32  zht467  阅读(142)  评论(0)  编辑  收藏  举报