POJ1149 PIGS 【最大流量】

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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

Sample Output

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;
}

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