POJ 1149 PIGS（最大流）

Description

Mirko works on a pig farm that consists of M locked pig-houses and Mirko can't unlock any pighouse because he doesn't have the keys. Customers come to the farm one after another. Each of them has keys to some pig-houses and wants to buy a certain number of pigs.  All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold.  More precisely, the procedure is as following: the customer arrives, opens all pig-houses to which he has the key, Mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses.  An unlimited number of pigs can be placed in every pig-house.  Write a program that will find the maximum number of pigs that he can sell on that day.

Input

The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of customers. Pig houses are numbered from 1 to M and customers are numbered from 1 to N.  The next line contains M integeres, for each pig-house initial number of pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000.  The next N lines contains records about the customers in the following form ( record about the i-th customer is written in the (i+2)-th line):  A K1 K2 ... KA B It means that this customer has key to the pig-houses marked with the numbers K1, K2, ..., KA (sorted nondecreasingly ) and that he wants to buy B pigs. Numbers A and B can be equal to 0.

Output

The first and only line of the output should contain the number of sold pigs.

题目大意：有n个人m个猪圈，每个猪圈里有一些猪，这n个人要买猪，每一个人会打开数个猪圈，买一些猪，然后这些猪圈的猪可以移动到另一些猪圈，问最多能卖多少猪。

思路：最大流。最朴素的想法是每个人建m个猪圈，不过这样搞点太多了。可以看到，若当两个猪圈被同时打开过，那么这此时两个猪圈就可以看成一个猪圈了。于是，对每一个猪圈，从源点到第一个打开它的人连一条边，容量为猪数，然后这个人再往下一个打开这个猪圈的人连一条边，容量为无穷大（其实这样两个人之间可能会有多条无穷大的边，不过我懒得搞，随便啦AC就好）。然后再从每个人连一条边到汇点，容量为这个人要买的猪的数量。最大流为答案。

PS：之前读入n、m的顺序搞错了WA了一次。n和m一样的样例太可恶了>_<

代码（16MS）：

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <queue>
 using namespace std;

 const int MAXN = ;
 const int MAXE = MAXN * MAXN;
 const int INF = 0x7fffffff;

 struct SAP {
     int head[MAXN], gap[MAXN], pre[MAXN], dis[MAXN], cur[MAXN];
     int next[MAXE], to[MAXE], flow[MAXE];
     int ecnt, n, st, ed;

     void init() {
         memset(head, , sizeof(head));
         ecnt = ;
     }

     void add_edge(int u, int v, int c) {
         to[ecnt] = v; flow[ecnt] = c; next[ecnt] = head[u]; head[u] = ecnt++;
         to[ecnt] = u; flow[ecnt] = ; next[ecnt] = head[v]; head[v] = ecnt++;
     }

     void bfs() {
         queue<int> que; que.push(ed);
         memset(dis, 0x3f, sizeof(dis));
         dis[ed] = ;
         while(!que.empty()) {
             int u = que.front(); que.pop();
             ++gap[dis[u]];
             for(int p = head[u]; p; p = next[p]) {
                 int &v = to[p];
                 if(flow[p ^ ] && dis[v] > n) {
                     dis[v] = dis[u] + ;
                     que.push(v);
                 }
             }
         }
     }

     int Max_flow(int ss, int tt, int nn) {
         st = ss, ed = tt, n = nn;
         int ans = , minFlow = INF, u;
         for(int i = ; i <= n; ++i) {
             cur[i] = head[i];
             gap[i] = ;
         }
         bfs();
         u = pre[st] = st;
         while(dis[st] < n) {
             bool flag = false;
             for(int &p = cur[u]; p; p = next[p]) {
                 int &v = to[p];
                 if(flow[p] && dis[u] == dis[v] + ) {
                     flag = true;
                     minFlow = min(minFlow, flow[p]);
                     pre[v] = u;
                     u = v;
                     if(u == ed) {
                         ans += minFlow;
                         while(u != st) {
                             u = pre[u];
                             flow[cur[u]] -= minFlow;
                             flow[cur[u] ^ ] += minFlow;
                         }
                         minFlow = INF;
                     }
                     break;
                 }
             }
             if(flag) continue;
             int minDis = n - ;
             for(int p = head[u]; p; p = next[p]) {
                 int &v = to[p];
                 if(flow[p] && dis[v] < minDis) {
                     minDis = dis[v];
                     cur[u] = p;
                 }
             }
             if(--gap[dis[u]] == ) break;
             ++gap[dis[u] = minDis + ];
             u = pre[u];
         }
         return ans;
     }
 } G;

 int n, m;
 int last[];
 int a[], b[];

 int main() {
     scanf("%d%d", &m, &n);
     G.init();
     int ss = n + , tt = n + ;
     for(int i = ; i <= m; ++i) scanf("%d", &a[i]);
     for(int i = ; i <= n; ++i) {
         int x, b;
         scanf("%d", &b);
         for(int j = ; j <= b; ++j) {
             scanf("%d", &x);
             if(last[x] == ) G.add_edge(ss, i, a[x]);
             else G.add_edge(last[x], i, INF);
             last[x] = i;
         }
         scanf("%d", &x); G.add_edge(i, tt, x);
     }
     printf("%d\n", G.Max_flow(ss, tt, tt));
 }