Poj1149 pigs [maximum flow]

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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

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

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Sample Output

7

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Source

Croatia OI 2002 Final Exam – First day

The main idea of the topic  Mirko Keep some pigs Pigs are kept in some pigsty The pigsty is locked He doesn't have a key himself （ Khan, ）  Only customers who want to buy pigs have keys  Customers come in turn Every customer will use his key to open some pigsty buy Walk some pigs Then lock it up.  Before locking Mirko There is a chance to allocate these opened pigpens again Pig of  Now give the number of pigs in each pigsty at the beginning All the keys of every customer And the number of pigs to buy ask Mirko How many pigs can you sell at most

Answer key ： For the first buyer of every pigsty , Add a source point to this person's side , The right is the number of pigs in this pigsty , For later people who want to buy the pigsty . Join a person who is the first to buy the pigsty to his side . take a temporary measure inf, Then join everyone to one side of the meeting point , The weight is the number of pigs that the person wants to buy . thus , The composition is finished .

#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; //  Here is the mark number num The first person to connect the pigsty 
                } else G[-pig[num]][i] = inf;
            }
            scanf("%d", &G[i][sink]);
        }
        printf("%d\n", Dinic());
    }
    return 0;
}

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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;
}
//  Source point , Confluence , The source point number must be the smallest , The sink number must be the largest 
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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