POJ1149 PIGS 【最大流量】

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16555		Accepted: 7416

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

题目大意

 Mirko养着一些猪 猪关在一些猪圈里面 猪圈是锁着的

他自己没有钥匙（汗）

 仅仅有要来买猪的顾客才有钥匙

 顾客依次来 每一个顾客会用他的钥匙打开一些猪圈 买

走一些猪 然后锁上

 在锁上之前 Mirko有机会又一次分配这几个已打开猪圈

的猪

 如今给出一開始每一个猪圈的猪数 每一个顾客全部的钥匙

和要买走的猪数 问Mirko最多能卖掉几头猪

题解：对于每一个猪圈的第一个购买的人，加入一条源点到这个人的边，权为这个猪圈的猪数，对于后来的且想要购买该猪圈的人。加入一条第一个购买该猪圈的人到该人的边。权为inf，然后加入每一个人到汇点一条边，权值为该人想要购买的猪的头数。至此，构图完毕。

#include <stdio.h>
#include <string.h>
#define inf 0x3fffffff
#define maxn 110
#define maxm 1002

int pig[maxm], m, n, sink;
int G[maxn][maxn], queue[maxn];
bool vis[maxn]; int Layer[maxn];

bool countLayer() {
memset(Layer, 0, sizeof(Layer));
int id = 0, front = 0, now, i;
Layer[0] = 1; queue[id++] = 0;
while(front < id) {
now = queue[front++];
for(i = 0; i <= sink; ++i)
if(G[now][i] && !Layer[i]) {
Layer[i] = Layer[now] + 1;
if(i == sink) return true;
else queue[id++] = i;
}
}
return false;
}

int Dinic() {
int minCut, pos, maxFlow = 0;
int i, id = 0, u, v, now;
while(countLayer()) {
memset(vis, 0, sizeof(vis));
vis[0] = 1; queue[id++] = 0;
while(id) {
now = queue[id - 1];
if(now == sink) {
minCut = inf;
for(i = 1; i < id; ++i) {
u = queue[i - 1];
v = queue[i];
if(G[u][v] < minCut) {
minCut = G[u][v];
pos = u;
}
}
maxFlow += minCut;
for(i = 1; i < id; ++i) {
u = queue[i - 1];
v = queue[i];
G[u][v] -= minCut;
G[v][u] += minCut;
}
while(queue[id - 1] != pos)
vis[queue[--id]] = 0;
} else {
for(i = 0; i <= sink; ++i) {
if(G[now][i] && Layer[now] + 1 == Layer[i] && !vis[i]) {
vis[i] = 1; queue[id++] = i; break;
}
}
if(i > sink) --id;
}
}
}
return maxFlow;
}

int main() {
//freopen("stdin.txt", "r", stdin);
int i, keys, num;
while(scanf("%d%d", &m, &n) == 2) {
sink = n + 1;
for(i = 1; i <= m; ++i)
scanf("%d", &pig[i]);
memset(G, 0, sizeof(G));
for(i = 1; i <= n; ++i) {
scanf("%d", &keys);
while(keys--) {
scanf("%d", &num);
if(pig[num] >= 0) {
G[0][i] += pig[num]; // 0 is source
pig[num] = -i; // 这里是标记第num个猪圈联通的第一个人
} else G[-pig[num]][i] = inf;
}
scanf("%d", &G[i][sink]);
}
printf("%d\n", Dinic());
}
return 0;
}

2015.4.20

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn = 105;
const int inf = 0x3f3f3f3f;
int G[maxn][maxn], M, N, S, T;
int pigHouse[maxn*10];

int Dinic(int s, int t);

void getMap()
{
memset(G, 0, sizeof(G));
S = 0; T = N + 1;

int i, j, K, pos;
for (i = 1; i <= M; ++i)
scanf("%d", &pigHouse[i]);

for (i = 1; i <= N; ++i) {
scanf("%d", &K);
while (K--) {
scanf("%d", &pos);
if (pigHouse[pos] >= 0) {
G[S][i] += pigHouse[pos];
pigHouse[pos] = -i;
} else {
G[-pigHouse[pos]][i] = inf;
}
}
scanf("%d", &G[i][T]);
}
}

void solve()
{
cout << Dinic(S, T) << endl;
}

int main()
{
while (cin >> M >> N) {
getMap();
solve();
}
return 0;
}

int queue[maxn];
bool vis[maxn]; int Layer[maxn];
bool countLayer(int s, int t) {
memset(Layer, 0, sizeof(Layer));
int id = 0, front = 0, now, i;
Layer[s] = 1; queue[id++] = s;
while(front < id) {
now = queue[front++];
for(i = s; i <= t; ++i)
if(G[now][i] && !Layer[i]) {
Layer[i] = Layer[now] + 1;
if(i == t) return true;
else queue[id++] = i;
}
}
return false;
}
// 源点，汇点，源点编号必须最小，汇点编号必须最大
int Dinic(int s, int t) {
int minCut, pos, maxFlow = 0;
int i, id = 0, u, v, now;
while(countLayer(s, t)) {
memset(vis, 0, sizeof(vis));
vis[s] = true; queue[id++] = s;
while(id) {
now = queue[id - 1];
if(now == t) {
minCut = inf;
for(i = 1; i < id; ++i) {
u = queue[i - 1];
v = queue[i];
if(G[u][v] < minCut) {
minCut = G[u][v];
pos = u;
}
}
maxFlow += minCut;
for(i = 1; i < id; ++i) {
u = queue[i - 1];
v = queue[i];
G[u][v] -= minCut;
G[v][u] += minCut;
}
while(queue[id - 1] != pos)
vis[queue[--id]] = false;
} else {
for(i = s; i <= t; ++i) {
if(G[now][i] && Layer[now] + 1 == Layer[i] && !vis[i]) {
vis[i] = 1; queue[id++] = i; break;
}
}
if(i > t) --id;
}
}
}
return maxFlow;
}