poj Minimum Cost 2516

Minimum Cost

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 16168		Accepted: 5672

Description

Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy has M supply places (marked from 1 to M), each provides K different kinds of goods (marked from 1 to K). Once shopkeepers order goods, Dearboy should arrange which supply place provide how much amount of goods to shopkeepers to cut down the total cost of transport. 

It's known that the cost to transport one unit goods for different kinds from different supply places to different shopkeepers may be different. Given each supply places' storage of K kinds of goods, N shopkeepers' order of K kinds of goods and the cost to transport goods for different kinds from different supply places to different shopkeepers, you should tell how to arrange the goods supply to minimize the total cost of transport.

Input

The input consists of multiple test cases. The first line of each test case contains three integers N, M, K (0 < N, M, K < 50), which are described above. The next N lines give the shopkeepers' orders, with each line containing K integers (there integers are belong to [0, 3]), which represents the amount of goods each shopkeeper needs. The next M lines give the supply places' storage, with each line containing K integers (there integers are also belong to [0, 3]), which represents the amount of goods stored in that supply place. 

Then come K integer matrices (each with the size N * M), the integer (this integer is belong to (0, 100)) at the i-th row, j-th column in the k-th matrix represents the cost to transport one unit of k-th goods from the j-th supply place to the i-th shopkeeper. 

The input is terminated with three "0"s. This test case should not be processed.

Output

For each test case, if Dearboy can satisfy all the needs of all the shopkeepers, print in one line an integer, which is the minimum cost; otherwise just output "-1".

Sample Input

1 3 3   
1 1 1
0 1 1
1 2 2
1 0 1
1 2 3
1 1 1
2 1 1

1 1 1
3
2
20

0 0 0

終於寫了一道最小費用最大流問題的題目了。很不容易啊！

題意：有N個店家、M個倉庫以及K種物品。接下來N行每行K個數，表示每個店家對每一種物品的需求。後面跟着M行，表示每個倉庫所存儲的每種物品的數目。

最後給出K個N*M的矩陣。對於第k個矩陣，它的第i行第j列表示——第j個倉庫運送第k種物品到第i個店家的費用。現在問你能否滿足所有店家的需求，若可以輸出最小的花費，反之輸出-1。

代碼:

#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;

const int Ma = 10052;
const int Me = 201;
const int inf = 202020202;
struct node
{
    int from, to, cap;
    int flow, cost, next;
}E[Ma];
int head[Me], top;
int dist[Me], pre[Me];
bool vis[Me], flag;
int source, sink;
int N, M, K;
int need[60][60], have[60][60];
int Sneed[60], Shave[60];
int used[60][60];

void init()
{
    top = 0;
    memset ( head, -1, sizeof(head) );
}

void AddEdge(int u, int v, int w, int c)
{
    node temp = {u, v, w, 0, c, head[u]};
    E[top] = temp;
    head[u] = top++;
    node temp1 = {v, u, 0, 0, -c, head[v]};
    E[top] = temp1;
    head[v] = top++;
}

bool SPFA(int s, int t)
{
    int i;
    queue<int> q;
    memset ( dist, inf, sizeof(dist) );
    memset ( vis, false, sizeof(vis) );
    memset ( pre, -1, sizeof(pre) );
    dist[s] = 0;
    vis[s] = true;
    q.push(s);
    while ( !q.empty() )
    {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for ( i = head[u];i != -1; i = E[i].next )
        {
            node temp = E[i];
            if ( dist[temp.to] > dist[u]+temp.cost&&temp.cap > temp.flow )
            {
                dist[temp.to] = dist[u]+temp.cost;
                pre[temp.to] = i;
                if ( !vis[temp.to] )
                {
                    vis[temp.to] = true;
                    q.push(temp.to);
                }
            }
        }
    }
    return pre[t] != -1;
}

void MCMF(int s, int t, int &cost)
{
    cost = 0;
    while(SPFA(s, t))
    {
        int Min = inf;
        for(int i = pre[t]; i != -1; i = pre[E[i^1].to])
        {
            node temp = E[i];
            Min = min(Min, temp.cap-temp.flow);
        }
        for(int i = pre[t]; i != -1; i = pre[E[i^1].to])
        {
            E[i].flow = E[i].flow+Min;
            E[i^1].flow = E[i^1].flow-Min;
            cost = cost+E[i].cost * Min;
        }
    }
}

void solve()
{
    int i, j, k;
    memset ( Sneed, 0, sizeof(Sneed) );
    memset ( Shave, 0, sizeof(Shave) );
    for ( i = 1;i <= N; i++ )
        for ( j = 1;j <= K; j++ )
        {
            scanf ( "%d", &need[i][j] );
            Sneed[j] = Sneed[j]+need[i][j];
        }
    for ( i = 1;i <= M; i++ )
        for ( j = 1;j <= K; j++ )
        {
            scanf ( "%d", &have[i][j] );
            Shave[j] = Shave[j]+have[i][j];
        }
    flag = true;
    for ( i = 1;i <= K; i++ )
        if ( Shave[i] < Sneed[i] )
        {
            flag = false;
            break;
        }
    int ans = 0;
    int cost = 0;
    for ( k = 1;k <= K; k++ )
    {
        for ( i = 1;i <= N; i++ )
        {
            for ( j = 1;j <= M; j++ )
                scanf ( "%d", &used[i][j] );
        }
        if ( !flag )
            continue;
        init();
        source = 0, sink = M+N+1;
        for ( i = 1;i <= N; i++ )
            AddEdge(i, sink, need[i][k], 0);
        for ( i = 1;i <= M; i++ )
            AddEdge(source, i+N, have[i][k], 0);
        for ( i = 1;i <= M; i++ )
            for ( j = 1;j <= N; j++ )
                AddEdge(i+N, j, inf, used[j][i]);
        MCMF(source, sink, cost);
        ans = ans+cost;
    }
    if ( flag )
        printf ( "%d\n", ans );
    else
        printf ( "-1\n" );
}
int main()
{
    while ( ~scanf ( "%d %d %d", &N, &M, &K )&&(N||M||K) )
    {
        solve();
    }
    return 0;
}

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