POJ2516 Minimum Cost

POJ2516 Minimum Cost

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 9392		Accepted: 3108

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

Sample Output

4
-1

****************************************************************

题目大意：有n个订单，m个仓库和k种物品。每个订单上有k种物品每种物品的需求，每个仓库里有k种物品的数量，还给出物品i从仓库j送到订单h所需要的花费。问满足订单要求的所需的最小花费。

解题思路：对每种物品分别计算。拆点+最小边权匹配。可怜我visy的位置写错，TLE了几次。

#include <stdio.h>
#include <string.h>
#include <vector>
#define N 55
#define INF 0x3f3f3f3f
using namespace std;

int tn[N][N],tm[N][N],tk[N][N][N];
int sum1[N],sum2[N];
int n,m,k,lack,limx,limy;
int gra[155][155],ix[1550],iy[155];
int lx[155],ly[155],mat[155];
int visx[155],visy[155];

int dfs(int x)
{
    visx[x]=1;
    for(int y=1;y<=limy;y++)
    {
        if(visx[mat[y]])continue;
        int t=lx[x]+ly[y]-gra[x][y];
        if(t==0)
        {
            visy[y]=1;
            if(!mat[y]||dfs(mat[y]))
            {
                mat[y]=x;
                return 1;
            }
        }
        else
            lack=min(lack,t);
    }
    return 0;
}

void re(void)
{
    memset(sum1,0,sizeof(sum1));
    memset(sum2,0,sizeof(sum2));
    for(int i=1;i<=n;i++)
        for(int j=1;j<=k;j++)
            scanf("%d",&tn[i][j]),sum1[j]+=tn[i][j];
    for(int i=1;i<=m;i++)
        for(int j=1;j<=k;j++)
            scanf("%d",&tm[i][j]),sum2[j]+=tm[i][j];
    for(int i=1;i<=k;i++)
        for(int j=1;j<=n;j++)
            for(int h=1;h<=m;h++)
                scanf("%d",&tk[i][j][h]);
}

void run(void)
{
    int ans=0;
    for(int h=1;h<=k;h++)
    {
        int id=0;
        limx=sum1[h];
        limy=sum2[h];
        if(limx>limy)
        {
            puts("-1");
            return ;
        }
        for(int i=1;i<=n;i++)
            for(int j=1;j<=tn[i][h];j++)
                ix[++id]=i;
        id=0;
        for(int i=1;i<=m;i++)
            for(int j=1;j<=tm[i][h];j++)
                iy[++id]=i;
        memset(lx,0,sizeof(lx));
        memset(ly,0,sizeof(ly));
        for(int i=1;i<=limx;i++)
            for(int j=1;j<=limy;j++)
            {
                gra[i][j]=1000-tk[h][ix[i]][iy[j]];
                lx[i]=max(lx[i],gra[i][j]);
            }
        memset(mat,0,sizeof(mat));
        for(int i=1;i<=limx;i++)
        {
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            lack=INF;
            while(!dfs(i))
            {
                for(int j=1;j<=limx;j++)
                    if(visx[j])
                    {
                        lx[j]-=lack;
                        visx[j]=0;
                    }
                for(int j=1;j<=limy;j++)
                    if(visy[j])
                    {
                        ly[j]+=lack;
                        visy[j]=0;
                    }
                lack=INF;
            }
        }
        for(int i=1;i<=limy;i++)
            if(mat[i])
                ans+=1000-gra[mat[i]][i];
    }
    printf("%d\n",ans);
}

int main()
{
    while(scanf("%d%d%d",&n,&m,&k)==3&&(n!=0||m!=0||k!=0))
    {
        re();
        run();
    }
    return 0;
}