标签：poj 网络最大流 dinic
PIGS
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 17598 |
|
Accepted: 7977 |
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
Source
Croatia OI 2002 Final Exam - First day
题目链接：http://poj.org/problem?id=1149
题目大意：m个猪圈，n个顾客，每个顾客有a把钥匙，k1,k2...ka，每个顾客想买b头猪，当每个顾客到来时，他将那些他拥有钥匙的猪圈全部打开，迈克从这些猪圈中挑出一些猪卖给他们，如果迈克愿意，迈克可以重新分配这些被打开的猪圈中的猪，当顾客离开时，猪圈将再次被锁上，注意：猪圈可容纳的猪的数量没有限制
题目分析：最大流的题，就是靠开脑洞建图：
1）单独设一个源点一个汇点
2）源点和每个猪圈的第一个顾客连边，边权是开始时猪圈中猪的数目
3）若源点和某个点之间有重边，则将权合并
4）顾客j紧跟在顾客i之后打开某个猪圈，则边<i,j>权为无穷大
5）每个顾客和汇点之间连边，边权为顾客希望买的猪的数目
建完图，闭眼Dinic，0ms
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
int const INF = 0x3fffffff;
int const MAXN = 105;
int const MAXM = 1005;
int cap[MAXN][MAXN], d[MAXN];
int pnum[MAXM], last[MAXM];
int m, n;
int src, sink;
bool BFS()
{
memset(d, -1, sizeof(d));
queue <int> q;
d[src] = 0;
q.push(src);
while(!q.empty())
{
int cur = q.front();
q.pop();
for(int i = 0; i <= sink; i++)
{
if(cap[cur][i] && d[i] == -1)
{
d[i] = d[cur] + 1;
q.push(i);
}
}
}
return d[sink] != -1;
}
int DFS(int t, int flow)
{
if(t == sink || flow == 0)
return flow;
int tmp = flow;
for(int i = 0; i <= sink; i++)
{
if(d[i] == d[t] + 1 && cap[t][i])
{
int mi = DFS(i, min(flow, cap[t][i]));
cap[t][i] -= mi;
cap[i][t] += mi;
flow -= mi;
}
}
return tmp - flow;
}
void Dinic()
{
int ans = 0;
while(BFS())
ans += DFS(src, INF);
printf("%d\n", ans);
}
int main()
{
memset(pnum, 0, sizeof(pnum));
memset(last, 0, sizeof(last));
scanf("%d %d", &m, &n);
src = 0;
sink = n + 1;
for(int i = 1; i <= m; i++)
scanf("%d", &pnum[i]);
for(int i = 1; i <= n; i++)
{
int num;
scanf("%d", &num);
for(int j = 0; j < num; j++)
{
int id;
scanf("%d", &id);
if(last[id] == 0)
cap[src][i] += pnum[id];
else
cap[last[id]][i] = INF;
last[id] = i;
}
scanf("%d", &cap[i][sink]);
}
Dinic();
}
POJ 1149 PIGS (网络最大流 Dinic 建对图你就赢了)
标签：poj 网络最大流 dinic
原文地址：http://blog.csdn.net/tc_to_top/article/details/45568577
