POJ 1149 PIGS (最大流Dinic)

题意：话说一个猪圈管理员，他本身没有猪圈的钥匙。每天会有许多顾客来买猪，这些顾客自己带着某些猪圈的钥匙。每当一个顾客来买猪，这些打开的猪圈里的猪可以随意流动，买完猪之后打开的猪圈全部关闭。现在已知每个猪圈里猪的的数量，每一名顾客拥有的钥匙以及他想购买的猪的数量。求管理员可以卖出的最大数量。

题解：构图是难点在于猪的流动。我是这样想的，假设顾客A可以打开了猪圈1,3,5，他需要购买numA头猪，由于每次被打开的猪圈之间的猪可以自由分配，那么猪圈1,3,5就可以视作一个集合X，当顾客A买完猪之后，集合X剩下的值为X-numA；如果下一次有一位顾客B可以打开猪圈2,4,5，他需要买numB头猪，将2,4,5其视作集合Y。由于X,Y之间有一个公共点5，那么集合Y的最大数量等于Y+X-numA。这样就比较清晰了。所有到达顾客A的流分作两部分，一部分流向汇点（numA）。另一部分流向顾客B（X-numA）。所以从A至B有一条边。

#include <iostream>using namespace std;#define MAX 2000#define INF 999999999#define min(a,b) (a<b?a:b)struct Edge{int st, ed;int next;int flow;} edge[MAX*10];bool key[101][1001]; // key[i][j] == true 表示第i名顾客拥有猪圈j的钥匙int head[MAX], out[MAX];int que[MAX], stk[MAX];int leve[MAX], house[MAX];int E, M, N, src, dest;void add_edge ( int u, int v, int val ){edge[E].st = u;edge[E].ed = v;edge[E].flow = val;edge[E].next = head[u];head[u] = E++;edge[E].st = v;edge[E].ed = u;edge[E].flow = 0;edge[E].next = head[v];head[v] = E++;}bool dinic_bfs (){int front, rear, u, v, i;memset(leve,-1,sizeof(leve));front = rear = 0;que[rear++] = src;leve[src] = 0;while ( front != ( rear + 1 ) % MAX ){u = que[front];front = ( front + 1 ) % MAX;for ( i = head[u]; i != -1; i = edge[i].next ){v = edge[i].ed;if ( leve[v] == -1 && edge[i].flow > 0 ){leve[v] = leve[u] + 1;que[rear] = v;rear = ( rear + 1 ) % MAX;}}}return leve[dest] >= 0;}int Dinic (){int top, u, i;int maxFlow = 0, minf;while ( dinic_bfs () ){top = 0; u = src;for ( i = src; i <= dest; i++ )              out[i] = head[i]; while ( out[src] != -1 ){if ( u == dest ){minf = INF;for ( i = top - 1; i >= 0; i-- )minf = min ( edge[stk[i]].flow, minf );maxFlow += minf;for ( i = top - 1; i >= 0; i-- ){edge[stk[i]].flow -= minf;edge[stk[i]^1].flow += minf;if ( edge[stk[i]].flow == 0 ) top = i;}u = edge[stk[top]].st;}else if ( out[u] != -1 && leve[u] + 1 == leve[edge[out[u]].ed] && edge[out[u]].flow > 0 ){stk[top++] = out[u];u = edge[out[u]].ed;}else{while ( top > 0 && u != src && out[u] == -1 )u = edge[stk[--top]].st;out[u] = edge[out[u]].next;}}}return maxFlow;}int main(){scanf("%d%d",&M,&N);src = E = 0;dest = M + N + 1;memset(key,0,sizeof(key));memset(head,-1,sizeof(head));int i, j, k, A, B;for ( i = 1; i <= M; i++ ){scanf("%d",&house[i]);add_edge ( src, i, house[i] );}for ( i = 1; i <= N; i++ ){scanf("%d",&A);for ( j = 1; j <= A; j++ ){scanf("%d",&k);key[i][k] = true;add_edge ( k, i+M, house[k] ); }for ( j = 1; j < i; j++ ) //枚举第i名顾客之前的顾客{for ( k = 1; k <= M; k++ )if ( key[j][k] && key[i][k] ) //若j,i有相同的钥匙，则可以连接一条从j到i的边，容量为无穷add_edge ( j+M, i+M, INF );}scanf("%d",&B);add_edge ( i+M, dest, B );}printf("%d\n",Dinic());return 0;}