poj 3422 Kaka's Matrix Travels 最小费最大流

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输入的时候没有取反，一直ole。
这里也是用到拆点，将一个点拆成p和q，这两个之间连接两条路，一条cap=1和cost=矩阵上的值，另一条为cap=k和cost=0。在将0和2*n *n+1看成源点和汇点。

#include<stdio.h>#include<string.h>#include<vector>#include<queue>#include<algorithm>using namespace std;const int N=10000+5;struct Edge{    int from,to,cap,flow,cost;};vector<Edge>edges;vector<int>G[N];int n,k,inq[N],d[N],p[N],a[N],c;void AddEdge(int from,int to ,int cap,int cost){    Edge tp;    tp.from=from,tp.to=to,tp.cap=cap,tp.flow=0,tp.cost=cost;    edges.push_back(tp);    tp.from=to,tp.to=from,tp.cap=0,tp.flow=0,tp.cost=-cost;    edges.push_back(tp);    int g=edges.size();    G[from].push_back(g-2);    G[to].push_back(g-1);}int BellmanFord(int s,int t,int &flow, int &cost){    int i,j,u;    for(i=0; i<=t; i++) d[i]=-1;    memset(inq,0,sizeof(inq));    d[s]=0;    inq[s]=1;    p[s]=0;    a[s]=0xfffffff;    queue<int>Q;    Q.push(s);    while(!Q.empty())    {        u=Q.front();        Q.pop();        inq[u]=0;        for(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]==-1) return 0;    flow+=a[t];    cost+=d[t]*a[t];    u=t;    while(u!=s)    {        edges[p[u]].flow+=a[t];        edges[p[u]^1].flow-=a[t];        u=edges[p[u]].from;    }    return 1;}int Mincost(int s,int t){    int flow=0,cost=0;    while(BellmanFord(s,t,flow,cost));    return cost;}int main(){    int i,j;    while(~scanf("%d%d",&n,&k))    {        for(i=0; i<=2*n*+1; i++) G[i].clear();        edges.clear();        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)            {                scanf("%d",&c);                int p=(i-1)*n+j;                int q=p+n*n;                AddEdge(p,q,1,c);                AddEdge(p,q,k,0);                if(i!=n) AddEdge(q,p+n,k,0);//下                if(j!=n) AddEdge(q,p+1,k,0);//右            }        AddEdge(0,1,k,0);        AddEdge(2*n*n,2*n*n+1,k,0);        int ans=Mincost(0,2*n*n+1);        printf("%d\n",ans);    }    return 0;}

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