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;
}