POJ 2516 Minimum Cost

问题：最小费用最大流

思路：针对对每一个物品建立网络流图求最小费用最大流，然后求和即可

PS：mincost忘了每次都初始化为0，导致一直WA！用尽了半天的时间去寻找，蓦然回首，原来mincost没有初始化！

#include <stdio.h>
#include <memory.h>

#define N 52
#define N2 104
#define M 5202
#define MAXVAL 0xfffffff
#define MOV(x) (x=(x+1)%N2)

int n,m,k,s,t;
int need[N][N],supply[N][N];
int nodevp[N2],nodeu[M],flow[M],cost[M],next[M],ind;

int getPartDataAndCheck()//首先得到need和supply的数据判断是否供大于求
{
    int i,j;

    scanf("%d %d %d",&n,&m,&k);
    if(0==n)    return 0;

    for(i=1;i<=n;i++)
        for(j=1;j<=k;j++)
            scanf("%d",need[i]+j);
    for(i=1;i<=m;i++)
        for(j=1;j<=k;j++)
            scanf("%d",supply[i]+j);
    for(i=1;i<=k;i++)
    {
        need[0][i]=0;
        for(j=1;j<=n;j++)
            need[0][i]+=need[j][i];
        supply[0][i]=0;
        for(j=1;j<=m;j++)
            supply[0][i]+=supply[j][i];

        if(need[0][i]>supply[0][i])    return -1;
    }
    
    return 1;
}

void eatData(int lineNum)//“吃数据”
{
//    char s[100];
//    while(lineNum--)    gets(s);

    char ch;
    while((ch=getchar())=='\n' || ch==' ');
    while(lineNum--)    while((ch=getchar())!='\n');

}

void addedge(int v,int u,int fw,int ct)
{
    nodeu[ind]=u;
    flow[ind]=fw;
    cost[ind]=ct;
    next[ind]=nodevp[v];
    nodevp[v]=ind++;
}

void buildGraph(int goodsNum)//以编号为goodsNum的物品，进行建图
{
    int i,j,val;

    s=0;    t=m+n+1;
    memset(nodevp,-1,sizeof(nodevp)); ind=0;
    for(i=1;i<=m;i++)
        if(supply[i][goodsNum])//去除不需要建的边
            addedge(s,i,supply[i][goodsNum],0) , addedge(i,s,0,0);
    for(i=1;i<=n;i++)
        if(need[i][goodsNum])//去除不需要建的边
            addedge(i+m,t,need[i][goodsNum],0) , addedge(t,i+m,0,0);
    for(i=1;i<=n;i++)
    {
        if(0==need[i][goodsNum])    eatData(1);//去除不需要建的边
        else
        {
            for(j=1;j<=m;j++)
            {
                scanf("%d",&val);
                if(supply[j][goodsNum])//去除不需要建的边
                    addedge(j,i+m,MAXVAL,val) , addedge(i+m,j,0,-val);
            }
        }
    }
                
}

int queue[N2],font,rear;
int inqueue[N2],pre[N2],dist[N2];
int SPFA()
{
    int i,v,u;
    for(i=1;i<=t;i++)    dist[i]=MAXVAL; dist[s]=0;
    memset(inqueue,0,sizeof(inqueue));    
    memset(pre,-1,sizeof(pre));
    font=rear=0;    queue[MOV(rear)]=s; inqueue[s]=1;

    while(font!=rear)
    {
        v=queue[MOV(font)];    inqueue[v]=0;
        for(i=nodevp[v];~i;i=next[i])
        {
            u=nodeu[i];
            if(flow[i] && dist[u]>dist[v]+cost[i])
            {
                dist[u]=dist[v]+cost[i]; pre[u]=i;
                if(!inqueue[u])    queue[MOV(rear)]=u,inqueue[u]=1;
            }
        }
    }
    
    if(dist[t]!=MAXVAL)    return 1;
    else    return 0;
}

int mincost_maxflow()
{
    int i,minflow,mincost=0;
    while(SPFA())
    {
        minflow=MAXVAL;
        for(i=pre[t];~i;i=pre[nodeu[i^1]])
            if(minflow>flow[i])
                minflow=flow[i];
        for(i=pre[t];~i;i=pre[nodeu[i^1]])
        {
            mincost+=cost[i]*minflow;
            flow[i]-=minflow;
            flow[i^1]+=minflow;
        }
    }
    
    return mincost;
}


void solve()
{
    int i,tmp,mincost; /* mincost=0; */
    while(tmp=getPartDataAndCheck())
    {
        if(tmp==-1)
        {
            eatData(k*n);
            printf("-1\n");
        }
        else
        {
            mincost=0;
            for(i=1;i<=k;i++)//针对每一个物品建图，求最小费用最大流!
            {
                buildGraph(i);
                mincost+=mincost_maxflow();
            }
            printf("%d\n",mincost);
        }
    }
}

int main()
{
//    freopen("input.txt","r",stdin);

    solve();

    return 0;
}