POJ1149 网络流最大流

[b]PIGS[/b]

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12372		Accepted: 5476

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.
Sample Input

3 3
3 1 10
2 1 2 2
2 1 3 3
1 2 6

Sample Output

7
题意：M个猪圈，N个顾客，每个顾客有一些的猪圈的钥匙，只能购买这些有钥匙的猪圈里的猪，而且要买一定数量的猪，每个猪圈有已知数量的猪，
但是猪圈可以重新打开，将猪的个数，重新分配，以达到卖出的猪的数量最多。

思路：刚学网络流，表示很菜很菜很菜~~
①构造网络，将顾客看成源点和汇点以外的结点，并设另外两个节点：源点和汇点。
②源点和每个猪圈的第一个顾客连边，边的权是开始时候猪圈中猪的数量。
③ 若源点和某个节点之间有重边，则将权合并
④顾客j紧跟顾客i之后打开某个猪圈，则<i.j>的权是正无穷。
⑤每个顾客和会点之间连边，边的权值是顾客所希望购买的猪的数量。
例如：样例中的就可以建立如图：

View Code

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
#define M 210
struct Matix{
int c,f;
}edge[M][M];
int n,m;
int s,t;
int resi[M][M];
int qu[M*M],qs,qe;
int pre[M];
int vis[M];
int maxflow,min_augment;
void find_augment_path(){
int i,v;
memset(vis,0,sizeof(vis));
qs=0;qu[qs]=s;
pre[s]=s;vis[s]=1;qe=1;
memset(resi,0,sizeof(resi));
memset(pre,0,sizeof(pre));
while(qs<qe&&pre[t]==0){
v=qu[qs++];
for(i=1;i<=n;i++){
if(vis[i]==0){
if(edge[v][i].c-edge[v][i].f>0){
resi[v][i]=edge[v][i].c-edge[v][i].f;
pre[i]=v;qu[qe++]=i;vis[i]=1;
}else if(edge[i][v].f>0){
resi[v][i]=edge[i][v].f;
pre[i]=v;qu[qe++]=i;vis[i]=1;
}
}
}
}
}
void augment_flow(){
int i=t,j;
if(pre[i]==0) {min_augment=0;return;}
j=0x7fffffff;
while(i!=s){
if(resi[pre[i]][i]<j) j=resi[pre[i]][i];
i=pre[i];
}
min_augment=j;
}
void update_flow(){
int i=t;
if(pre[i]==0) return ;
while(i!=s){
if(edge[pre[i]][i].c-edge[pre[i]][i].f>0)
edge[pre[i]][i].f+=min_augment;
else if(edge[i][pre[i]].f>0) edge[pre[i]][i].f+=min_augment;
i=pre[i];
}
}
void solve(){
s=1;t=n;
maxflow=0;
while(1){
find_augment_path();
augment_flow();
maxflow+=min_augment;
if(min_augment>0) update_flow();
else return ;
}
}
int main(){
int i;
int u,v,w;
while(scanf("%d%d",&m,&n)!=EOF){
memset(edge,0,sizeof(edge));
for(i=0;i<m;i++){
scanf("%d%d%d",&u,&v,&w);
edge[u][v].c+=w;
}
solve();
printf("%d\n",maxflow);
}
return 0;
}