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
;
int head
,dis
,route
;

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\n",-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];
}
}