hdu 2686 最小费用最大流

#include "stdio.h" //hdu 2686 最小费用最大流
#include "string.h"
#include "queue"
using namespace std;

#define N 2005
#define INF 0x3fffffff

struct node {
	int u,v,w,k;
	int next;
}edge[4*N*10];

int n,ans,idx;
bool mark[N];
int head[N],dis[N],route[N];

void init();
int SPFA(int start,int end);
void EK(int start,int end);
void adde(int u,int v,int w,int k);
void addedge(int u,int v,int w,int k);

int main()
{
	int i,j;
	int w,k,u,v;
	while(scanf("%d",&n)!=-1)
	{
		init();
		k = n*n;
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				u = j+(i-1)*n;
				scanf("%d",&w);
				adde(u,u+k,-w,1);  //因为求最大费用流，故取反
				if(j!=n)	adde(u+k,u+1,0,INF);
				if(i!=n)	adde(u+k,u+n,0,INF);
			}
		}
		int start=0,end = 2*n*n+1;
		adde(start,1,0,2);
		adde(1,k+1,0,1);  //为1->n*n再添一条边，花费为0
		adde(2*k,end,0,2);
		adde(k,2*k,0,1); //为n*n->2*n*n再添一条边，花费为0
		while(SPFA(start,end))
			EK(start,end);
		printf("%d
",-ans);
	}
	return 0;
}

void init()   //初始化
{
	ans = 0;
	idx = 0;
	memset(head,-1,sizeof(head));
}

void adde(int u,int v,int w,int k)
{
	addedge(u,v,w,k);
	addedge(v,u,-w,0);
}

void addedge(int u,int v,int w,int k)
{
	edge[idx].u = u;
	edge[idx].v = v;
	edge[idx].w = w;
	edge[idx].k = k;
	edge[idx].next = head[u];
	head[u] = idx;
	idx++;
}

int SPFA(int start ,int end)   //自己的理解，带负权的图中，能用SPFA算法，是因为本题中没有带负权的环！
{
	int i;
	for(i=0;i<=end;i++) dis[i] = INF;
	memset(mark,false,sizeof(mark));
	memset(route,-1,sizeof(route));
	dis[start] = 0;
	queue<int> q;
	q.push(start);
	mark[start] = true;
	int x,y;
	while(!q.empty())
	{
		x = q.front();
		q.pop();
		for(i=head[x];i!=-1;i = edge[i].next)
		{ 
			y = edge[i].v;
			if(edge[i].k && dis[y] > dis[x] + edge[i].w)
			{
				dis[y] = dis[x] + edge[i].w;
				route[y] = i;
				if(mark[y] == false)
				{
					mark[y] = true;
					q.push(y);
				}
			}
		}
		mark[x] = false;
	}
	if(route[end]==-1) return 0;
	return 1;
}

void EK(int start,int end)  
{
	int x,y=route[end];

	while(y!=-1)
	{
		x = y^1;
		edge[y].k--;
		edge[x].k++;
		ans += edge[y].w*s;
		y = route[edge[y].u];
	}
}