Minimum Cost
Time Limit: 4000MS | Memory Limit: 65536K | |
Total Submissions: 11530 | Accepted: 3892 |
Description
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
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
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 題意:n個店,m個供貨點,k種貨物。一個n*k的矩陣代表第i個店主對第j種貨物的需求,接下來的m*k行,表示第i個供貨點第j種的貨物。在接下來的k個n*m個矩陣表示,第j個供貨點供給第i個店的第k種貨物的花費。求最小花費。
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define INF 9999999
#define maxn 205
int lx[maxn],ly[maxn];
int w[maxn][maxn];
int Match[maxn],visx[maxn],visy[maxn];
int slack[maxn];
int ans,n,m,k;
int need[maxn][maxn],have[maxn][maxn],cost[maxn][maxn][maxn];
int indexA[maxn],indexB[maxn];
int A,B;
bool findPath(int x)
{
int tmp;
visx[x]=1;
for(int y=1; y<=B; y++)
{
if(visy[y])continue;
if(w[x][y]==lx[x]+ly[y])
{
visy[y]=1;
if(!Match[y]||findPath(Match[y]))
{
Match[y]=x;
return true;
}
}
else
slack[y]=min(slack[y],w[x][y]-lx[x]-ly[y]);
}
return false;
}
void km()
{
memset(Match,0,sizeof(Match));
memset(ly,0,sizeof(ly));
for(int i=1; i<=A; i++)
{
lx[i]=INF;
for(int j=1; j<=B; j++)
lx[i]=min(lx[i],w[i][j]);
}
for(int x=1; x<=A; x++)
{
for(int i=1; i<=B; i++)
slack[i]=INF;
while(true)
{
memset(visx,0,sizeof(visx));
memset(visy,0,sizeof(visy));
if(findPath(x))break;
int tmp=INF;
for(int i=1; i<=B; i++)
{
if(!visy[i])
{
if(tmp>slack[i])
tmp=slack[i];
}
}
if(tmp==INF)return ;
for(int i=1; i<=A; i++)
if(visx[i])lx[i]+=tmp;
for(int i=1; i<=B; i++)
if(visy[i])ly[i]-=tmp;
}
}
}
int main()
{
int cas;
while(scanf("%d%d%d",&n,&m,&k))
{
if(n==0&&m==0&&k==0)break;
int a,b,c;
ans=0;
for(int i=1; i<=n; i++)
for(int j=1; j<=k; j++)
{
cin>>need[i][j];
}
for(int i=1; i<=m; i++)
for(int j=1; j<=k; j++)
{
cin>>have[i][j];
}
for(int i=1; i<=k; i++)
for(int j=1; j<=n; j++)
for(int l=1; l<=m; l++)
scanf("%d",&cost[i][j][l]);
int flag1=0;
for(int i=1; i<=k; i++)
{
int sumNeed=0,sumHave=0;
for(int j=1; j<=n; j++)
sumNeed+=need[j][i];
for(int j=1; j<=m; j++)
sumHave+=have[j][i];
if(sumNeed>sumHave)
{
flag1=1;
cout<<"-1"<<endl;
break;
}
}
if(flag1)continue;
for(int i=1; i<=k; i++)
{
A=0;
B=0;
for(int j=1; j<=n; j++)
for(int l=1; l<=need[j][i]; l++)
indexA[++A]=j;
for(int j=1; j<=m; j++)
for(int l=1; l<=have[j][i]; l++)
indexB[++B]=j;
for(int j=1; j<=A; j++)
for(int l=1; l<=B; l++)
w[j][l]=cost[i][indexA[j]][indexB[l]];
km();
//int flag=0;
for(int i=1; i<=B; i++)
{
if(Match[i])
ans+=w[Match[i]][i];
}
}
cout<<ans<<endl;
}
return 0;
}