POJ 2516 Minimum Cost (费用流)

题面

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

题解

我首先翻译一下题目

尤其说清楚一下输入的格式。。网上的说的都不太清楚

第一行是三个数字:N,M,K

分别表示有N个商店，M个供货商，K中货物

接下来是N行，每行K个整数

对于第i行第j列，表示的是第i个商店对于货物j的需求

再接着，M行，每行K个整数

对于第i行第j列，表示的是第i个供货商对于货物j的存货

接下来有K个N*M的矩形

第X个矩形的第i行第j列表示的是

从供货商j 运送一个单位的 货物X 到商店i的 花费

最后要求的是

在满足 所有商店的供应的 情况下的 最小花费

如果无法满足，则输出-1

题解：

首先弄清楚题目的意思

考虑k种货物都是独立的，因此只需要考虑k遍最小的费用然后求和

对于每一次的最小费用，显然直接求解最小费用流即可，

对于图的构建并不难，

但是要考虑清楚每条边的容量

从汇点向每个供货商连接一条容量为存货数量，费用为0的边

然后从每个供货商向每个商店连接一条容量为INF，费用为花费的边（容量连接成存货数量也行）

从商店向汇点连接一条容量为需求，费用为0的边

求解K次最小费用流累加答案即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
#define MAX 200
#define MAXL 200000
#define INF 1000000000
struct Line
{
	int v,next,w,fb,fy;
}e[MAXL];
int h[MAX],cnt=1,cost,ff;
int tot[MAX],need[MAX][MAX],have[MAX][MAX],Cost[MAX][MAX][MAX];
int pe[MAX],pr[MAX],Ans;
inline void Add(int u,int v,int w,int fy)
{
	e[cnt]=(Line){v,h[u],w,cnt+1,fy};
	h[u]=cnt++;
	e[cnt]=(Line){u,h[v],0,cnt-1,-fy};
	h[v]=cnt++;
}
int dis[MAX],S,T,N,M,K;

bool vis[MAX];
bool SPFA()
{
	memset(dis,63,sizeof(dis));
	dis[S]=0;
	queue<int> Q;while(!Q.empty())Q.pop();
	Q.push(S);
	memset(vis,0,sizeof(vis));
	while(!Q.empty())
	{
		int u=Q.front();Q.pop();
		vis[u]=false;
		for(int i=h[u];i;i=e[i].next)
		{
			int f=dis[u]+e[i].fy,v=e[i].v;
			if(e[i].w&&dis[v]>f)
			{
				dis[v]=f;
				pe[v]=i;
				pr[v]=u;
				if(!vis[v])
				{
					vis[v]=true;
					Q.push(v);
				}
			}
		}
	}
	if(dis[T]==dis[T+1])return false;//增广失败
	int re=INF;
	for(int v=T;v!=S;v=pr[v])
		re=min(re,e[pe[v]].w);//计算增广的最大流
	for(int v=T;v!=S;v=pr[v])
	{
		e[pe[v]].w-=re;
		e[e[pe[v]].fb].w+=re;
    }
	ff+=re;
	cost+=re*dis[T];
	return true;
}
int main()
{
	while(233)
	{
		cin>>N>>M>>K;
		if(N==0&&M==0&&K==0)break;
		memset(tot,0,sizeof(tot));
		for(int i=1;i<=N;++i)
		{
			for(int j=1;j<=K;++j)
			{
				cin>>need[i][j];
				tot[j]+=need[i][j];
			}
		}
		for(int i=1;i<=M;++i)
		{
			for(int j=1;j<=K;++j)
			{
				cin>>have[i][j];
				tot[j]-=have[i][j];
			}
		}
		for(int i=1;i<=K;++i)
		{
			for(int j=1;j<=N;++j)
				for(int k=1;k<=M;++k)
					cin>>Cost[i][j][k];
		}
		S=0;T=N+M+1;
		bool fl=true;
		for(int k=1;k<=K;++k)
			if(tot[k]>0)//需求多于提供
			{
				fl=false;
				break;
			}
		if(!fl)
		{
			fl=true;
			printf("%d\n",-1);
			continue;
		}
		Ans=0;
		for(int k=1;k<=K;++k)//K遍费用流
		{
			cnt=1;
			memset(h,0,sizeof(h));
			for(int j=1;j<=M;++j)
			{
			 	Add(S,j,have[j][k],0);
				for(int i=1;i<=N;++i)
					Add(j,i+M,have[j][k],Cost[k][i][j]);
			}
			for(int i=1;i<=N;++i)
				Add(i+M,T,need[i][k],0);
			cost=ff=0;
			while(SPFA());
			Ans+=cost;
		}
		printf("%d\n",Ans);
	}
	return 0;
}