UVa10779 Collectors' Problem（ 网络流，建图）

这是一道建图的题，只要能想清楚怎么建图，就没有问题

建图：Bob和所有粘贴相连，边的容量是他所拥有粘贴的个数；

            其他人和他们所拥有的粘贴相连，边容量为手中物品个数减1，因为他们要保留一个

            其他人没有的粘贴，从粘贴连到这个人，容量为1，他只能接受一个这种粘贴

            每种粘贴和汇点相连，容量为1

代码如下：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int inf = 99999999;const int Maxn = 40;int S, E, T, n, m;int map[Maxn][Maxn], flow[Maxn][Maxn], a[Maxn], p[Maxn];void buildMap() {    int ki, t;    for ( int i = 1; i <= n; ++i ) {        scanf("%d", &ki);        for ( int j = 0; j < ki; ++j ) {              scanf("%d", &t);            t += n;             map[i][t]++;        }        if ( i > 1 ) for ( int j = n + 1; j <= n + m; ++j )             if ( map[i][j] ) map[i][j]--;            else if ( map[i][j] == 0 ) map[j][i]++;    }    for ( int i = n + 1; i <= n + m; ++i ) map[i][E]++; //将表示物品的点和汇点相连，容量为1}int maxFlow() {    queue < int > q; int f = 0;    memset( flow, 0, sizeof(flow) );    while(1) {        memset( a, 0, sizeof(a) );        a[S] = inf;        q.push( S );        while( !q.empty() ) {            int u = q.front(); q.pop();            for( int v = 1; v <= m + n + 1; ++v ) {                if ( !a[v] && map[u][v] > flow[u][v] ) {                    p[v] = u; q.push(v);                    a[v] = min(a[u], map[u][v] - flow[u][v]);                }            }        }        if ( a[E] == 0 ) break;        for ( int u = E; u != S; u = p[u] ) {            flow[p[u]][u] += a[E];            flow[u][p[u]] -= a[E];        }        f += a[E];    }    return f;}int main(){    int icase = 1;    while ( scanf("%d", &T) != EOF ) {        while ( T-- ) {            scanf("%d%d", &n, &m);            memset( map, 0, sizeof(map) );            S = 1, E = m + n + 1;            buildMap();            printf("Case #%d: %d\n", icase++, maxFlow());        }    }}