poj 3422 (费用流)

从左上角到有下角k次能获得的最大值。

跟hdu 2686一样的题目，这题一个点可以重复走，只能得到一次值。

#include<stdio.h>
#include<string.h>
#include<queue>
const int N=5100;
const int inf=0x3fffffff;
using namespace std;
int dist[N],head[N],num,start,end,n,vis[N],pre[N];
struct edge
{
	int st,ed,cost,flow,next;
}e[N*10];
void addedge(int x,int y,int c,int w)
{
	e[num].st=x;e[num].ed=y;e[num].cost=c; e[num].flow=w;e[num].next=head[x];head[x]=num++;
	e[num].st=y;e[num].ed=x;e[num].cost=-c;e[num].flow=0;e[num].next=head[y];head[y]=num++;
}
int SPFA()
{
	queue<int>Q;
	int i,v,u;
	for(i=0;i<=end;i++)
	{dist[i]=-1;vis[i]=0;pre[i]=-1;}
	dist[start]=0;vis[start]=1;
	Q.push(start);
	while(!Q.empty())
	{
		u=Q.front();Q.pop();
		vis[u]=0;
		for(i=head[u];i!=-1;i=e[i].next)
		{
			v=e[i].ed;
			if(e[i].flow>0&&dist[v]<dist[u]+e[i].cost)
			{
				dist[v]=dist[u]+e[i].cost;
				pre[v]=i;
				if(vis[v]==0)
				{
					Q.push(v);
					vis[v]=1;
				}
			}
		}
	}
	if(pre[end]==-1)
		return 0;
	return 1;
}
int Maxcost()
{
	int i,maxflow=0,minflow,maxcost=0;
	while(SPFA())
	{
		minflow=inf;
		for(i=pre[end];i!=-1;i=pre[e[i].st])
		 if(minflow>e[i].flow)
			 minflow=e[i].flow;
		 maxflow+=minflow;
		 for(i=pre[end];i!=-1;i=pre[e[i].st])
		 {
			 e[i].flow-=minflow;
			 e[i^1].flow+=minflow;
			 maxcost+=e[i].cost;
		 }
	}
	//printf("maxflow=%d\n",maxflow);
	return maxcost;
}
int main()
{
	int i,j,t,x,w,k;
	while(scanf("%d%d",&n,&k)!=-1)
	{
		t=n*n;start=0;end=t*2+1;num=0;
		memset(head,-1,sizeof(head));
		addedge(start,1,0,k);
		addedge(t+t,end,0,k);
		for(i=1;i<=n;i++)
			for(j=1;j<=n;j++)
			{
				scanf("%d",&w);
				x=i*n+j-n;
				addedge(x,x+t,w,1);
				addedge(x,x+t,0,k-1);
				if(j+1<=n)
					addedge(x+t,x+1,0,k);
				if(i+1<=n)
					addedge(x+t,x+n,0,k);
			}
			printf("%d\n",Maxcost());
	}
	return 0;
}