POJ 1149 PIGS

/* Tag

admonds karp求网络最大流

这题建图的方法很重要：

1）对于client i，所有i先到达的猪圈的猪的数量和作为0和i之间的权值

2）对于client i，如果其可到的猪圈k已经被另一个client j先到给占了，那么将j 到 i 的边的权值设置为无穷大

3）对于client i，i 到 clientNum + 1（终点）的权值设为i可以买的最多的猪的数量

4）其他边权值均设置为0

*/

#include <iostream> #include <queue> #define MAX_N 102 #define MAX_M 1000 #define MAX_VAL 1000000 using namespace std; int graph[MAX_N + 1][MAX_N + 1]; int roomP[MAX_M + 1]; int cNum, rNum; int gf[MAX_N + 1][MAX_N + 1]; int f[MAX_N + 1][MAX_N + 1]; queue<int> q; bool v[MAX_N + 1]; int pre[MAX_N + 1]; int head, tail; //寻找增广路径 int getAP() { int i; for(i = 1; i <= cNum + 1; i++) { pre[i] = -1; v[i] = false; } pre[0] = -1; v[0] = true; while(!q.empty()) q.pop(); q.push(0); while(!q.empty()) { int curN = q.front(); q.pop(); for(i = 0; i <= cNum + 1; i++) { if(i != curN && gf[curN][i] > 0 && !v[i]) { v[i] = true; pre[i] = curN; q.push(i); } } } if(pre[cNum + 1] == -1) return -1; else { int curPos = cNum + 1; int res = INT_MAX; int prePos = pre[curPos]; while(prePos != -1) { if(gf[prePos][curPos] < res) res = gf[prePos][curPos]; curPos = prePos; prePos = pre[curPos]; } return res; } } void edmondsKarp() { int minW = 0; while((minW = getAP()) != -1) { int curPos = cNum + 1; while(pre[curPos] != -1) { int prePos = pre[curPos]; gf[prePos][curPos] -= minW; gf[curPos][prePos] += minW; f[prePos][curPos] += minW; f[curPos][prePos] = -f[prePos][curPos]; curPos = prePos; } } } int main() { int i, j, keyN, keyId, upQ; cin>>rNum>>cNum; for(i = 1; i <= rNum; i++) cin>>roomP[i]; for(i = 1; i <= cNum; i++) { cin>>keyN; for(j = 1; j <= keyN; j++) { cin>>keyId; if(roomP[keyId] >= 0) { graph[0][i] += roomP[keyId]; gf[0][i] = graph[0][i]; roomP[keyId] = -i; } else { graph[-roomP[keyId]][i] = MAX_VAL; gf[-roomP[keyId]][i] = graph[-roomP[keyId]][i]; } } cin>>upQ; graph[i][cNum + 1] = upQ; gf[i][cNum + 1] = upQ; } edmondsKarp(); int res = 0; for(i = 0; i <= cNum + 1; i++) { if(i != 0 && graph[0][i] > 0) res += f[0][i]; } cout<<res<<endl; return 0; }