poj 1273.PIG 解题报告

/*
      最大流SAP
      邻接表
      思路：基本源于FF方法，给每个顶点设定层次标号，和允许弧。
      优化:
      1、当前弧优化（重要）。
      1、每找到以条增广路回退到断点（常数优化）。
      2、层次出现断层，无法得到新流（重要）。
      时间复杂度（m*n^2）
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#define ms(a,b) memset(a,b,sizeof a)
using namespace std;
const int INF = 500;
int G[INF][INF];
struct node {
    int v, c, next;
} edge[INF*INF*4];
int  pHead[INF*INF], SS, ST, nCnt;
void addEdge (int u, int v, int c) {
    edge[++nCnt].v = v; edge[nCnt].c = c, edge[nCnt].next = pHead[u]; pHead[u] = nCnt;
    edge[++nCnt].v = u; edge[nCnt].c = 0, edge[nCnt].next = pHead[v]; pHead[v] = nCnt;
}
int SAP (int pStart, int pEnd, int N) {
    int numh[INF], h[INF], curEdge[INF], pre[INF];
    int cur_flow, flow_ans = 0, u, neck, i, tmp;
    ms (h, 0); ms (numh, 0); ms (pre, -1);
    for (i = 0; i <= N; i++) curEdge[i] = pHead[i];
    numh[0] = N;
    u = pStart;
    while (h[pStart] <= N) {
        if (u == pEnd) {
            cur_flow = 1e9;
            for (i = pStart; i != pEnd; i = edge[curEdge[i]].v)
                if (cur_flow > edge[curEdge[i]].c) neck = i, cur_flow = edge[curEdge[i]].c;
            for (i = pStart; i != pEnd; i = edge[curEdge[i]].v) {
                tmp = curEdge[i];
                edge[tmp].c -= cur_flow, edge[tmp ^ 1].c += cur_flow;
            }
            flow_ans += cur_flow;
            u = neck;
        }
        for ( i = curEdge[u]; i != 0; i = edge[i].next)
            if (edge[i].c && h[u] == h[edge[i].v] + 1)     break;
        if (i != 0) {
            curEdge[u] = i, pre[edge[i].v] = u;
            u = edge[i].v;
        }
        else {
            if (0 == --numh[h[u]]) continue;
            curEdge[u] = pHead[u];
            for (tmp = N, i = pHead[u]; i != 0; i = edge[i].next)
                if (edge[i].c)  tmp = min (tmp, h[edge[i].v]);
            h[u] = tmp + 1;
            ++numh[h[u]];
            if (u != pStart) u = pre[u];
        }
    }
    return flow_ans;
}
/*
       poj1149 最大流
       建图：
       每个顾客为一个节点，从源点到每个猪圈的第一个顾客连接一条容量为猪圈猪数目的边
       从第一个顾客到同一个猪圈的其它顾客连一条容量无限的边
       每个顾客到汇点的边的容量为他最大买的数量

*/
int k, c, m, n,x,y,z;
int vis[INF],sum[INF];
void solve(){
       nCnt=1;
       for(int i=1;i<=ST;i++)
       for(int j=1;j<=ST;j++){
              if(G[i][j]) addEdge(i,j,G[i][j]);
       }
       int ans=SAP(SS,ST,ST);
       printf("%d\n",ans);
}
int main() {
    /*
           先用邻接矩阵统计容量，再用前向星存边，表头在pHead[],初始化nCnt=1
           SS,ST分别为源点和汇点
    */
       scanf("%d %d",&m,&n);
       SS=n+1,ST=n+2;
       for(int i=1;i<=m;i++)
              scanf("%d",&sum[i]);
       for(int i=1;i<=n;i++){
              scanf("%d",&k);
              for(int j=1;j<=k;j++){
                     scanf("%d",&x);
                     if(!vis[x]){
                        G[SS][i]+=sum[x];
                        vis[x]=i;
                     }
                     else{
                        G[vis[x]][i]=0xffff;
                     }
              }
              scanf("%d",&k);
              G[i][ST]=k;
       }
       solve();
    return 0;
}