POJ 2516Minimum Cost(最小费用最大流)
题意：给出n个客户对k种商品的需求量，又给出m个仓库对k种物品的存货量以及对k种物品从i仓库到j客户的一个物品的运费价格，让判断是否可以满足客户需求，然后就是如果满足求出最小的运费.
思路：简单的最小费用最大流，注意建图细节即可。下面解释转自网上。
首先必须判断m个仓库是否有足够的k种商品给n个客户,如果不足,那么明显就是不行的. 下面假设仓库的商品足够的话:
对于每一种商品我们都算出满足满足顾客需求量的最小运费即可.所以我们对K种商品分开处理如下,假设当前处理第x种商品,建图如下:
源点s编号0, m个仓库编号1到m, n个顾客编号m+1到m+n, 汇点编号m+n+1.
从源点s到每个仓库i有边(s, i, 仓库i对商品x的存货量, 0)
从每个仓库i到顾客j有边(i, j, INF, 仓库i到顾客j的单位X商品的运费)
从每个顾客j到汇点t有边(j, t, 顾客j对X商品的需求量, 0)
然后我们求最小费用最大流即可求出N个顾客对第X种商品的最小运费.
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 300
#define LL long long
int cas=1,T;
const int INF = 1e9;
struct Edge
{
int from,to,cap,flow,cost;
Edge(){}
Edge(int u,int v,int c,int f,int co):from(u),to(v),cap(c),flow(f),cost(co){}
};
int n,m;
struct MCMF
{
// int s,t;
vector<Edge>edges;
vector<int>G[maxn];
int inq[maxn]; //是否在队列中
int d[maxn];
int p[maxn]; //上一条弧
int a[maxn]; //可改进量
void init()
{
for (int i = 0;i<=n+m+1;i++)
G[i].clear();
edges.clear();
}
void AddEdge(int from,int to,int cap,int cost)
{
edges.push_back(Edge(from,to,cap,0,cost));
edges.push_back(Edge(to,from,0,0,-cost));
int mm = edges.size();
G[from].push_back(mm-2);
G[to].push_back(mm-1);
}
bool BellmanFord(int s,int t,int &flow,int &cost)
{
for (int i = 0;i<=n+m+1;i++)
d[i]=INF;
memset(inq,0,sizeof(inq));
d[s]=0;
inq[s]=1;
p[s]=0;
a[s]=INF;
queue<int>q;
q.push(s);
while (!q.empty())
{
int u = q.front();q.pop();
inq[u]=0;
for (int i = 0;i<G[u].size();i++)
{
Edge &e = edges[G[u][i]];
if (e.cap > e.flow && d[e.to]>d[u]+e.cost)
{
d[e.to]=d[u]+e.cost;
p[e.to]=G[u][i];
a[e.to]=min(a[u],e.cap-e.flow);
if (!inq[e.to])
{
q.push(e.to);
inq[e.to]=1;
}
}
}
}
if (d[t]==INF)
return false; //s-t不连通，失败退出
flow+=a[t];
cost+=d[t]*a[t];
int u = t;
while (u!=s)
{
edges[p[u]].flow+=a[t];
edges[p[u]^1].flow-=a[t];
u=edges[p[u]].from;
}
return true;
}
int Mincost(int s,int t)
{
int flow = 0;
int cost = 0;
while (BellmanFord(s,t,flow,cost));
return cost;
}
}mc;
int need[60][60]; //i顾客对j商品的需求量
int have[60][60]; //i仓库对j商品的供应量
int cost[60][60][60]; //仓库j到i顾客对x商品的运费
int main()
{
//freopen("in","r",stdin);
int k;
while (scanf("%d%d%d",&n,&m,&k)!=EOF && n)
{
int goods[maxn];
int flag = 1;
memset(goods,0,sizeof(goods));
for (int i = 1;i<=n;i++)
for (int j = 1;j<=k;j++)
{
scanf("%d",&need[i][j]);
goods[j]+=need[i][j];
}
for (int i = 1;i<=m;i++)
for (int j = 1;j<=k;j++)
{
scanf("%d",&have[i][j]);
goods[j]-=have[i][j];
}
for (int h = 1;h<=k;h++)
for (int i = 1;i<=n;i++)
for (int j = 1;j<=m;j++)
scanf("%d",&cost[h][i][j]);
for (int i = 1;i<=k;i++)
if (goods[i]>0)
{
flag = 0;
break;
}
if (!flag)
{
printf("-1\n");
continue;
}
int mins = 0;
// mc.init();
for(int g = 1;g<=k;g++)
{
mc.init();
for (int i = 1;i<=m;i++)
mc.AddEdge(0,i,have[i][g],0);
for (int i = 1;i<=n;i++)
mc.AddEdge(m+i,m+n+1,need[i][g],0);
for (int i = 1;i<=m;i++)
for (int j = 1;j<=n;j++)
mc.AddEdge(i,j+m,INF,cost[g][j][i]);
mins+=mc.Mincost(0,n+m+1);
}
printf("%d\n",mins);
}
//scanf("%d",&T);
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
}