http://poj.org/problem?id=2135
题意:就是让你从1->n->1,不能走重复的边,问你最小时间(花费);
题解: 抽象源点与汇点,源点连1,费用为0,容量为2,汇点连n,费用为0,容量为2,中间的边的容量为1,权值为花费,跑最大流,限制满流情况为2就好啦,
代码如下:
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define maxn 2000
#define maxm 222222
#define INF 0x3f3f3f3f
int head[maxn],d[maxn],s,e,no,dis[maxn][maxn],vis[maxn],pre[maxn];
struct point
{
int u,v,flow,next,cost;
point(){};
point(int x,int y,int z,int w,int c):u(x),v(y),next(z),flow(w),cost(c){};
}p[maxm];
void add(int x,int y,int z,int c)
{
p[no]=point(x,y,head[x],z,c);
head[x]=no++;
p[no]=point(y,x,head[y],0,-c);
head[y]=no++;
}
void init()
{
memset(head,-1,sizeof(head));
no=0;
}
bool spfa()
{
int i,x,y;
queue<int>q;
memset(d,0x3f,sizeof(d));
memset(vis,false,sizeof(vis));
memset(pre,-1,sizeof(pre));
d[s]=0;
vis[s]=true;
q.push(s);
while(!q.empty())
{
x=q.front();
q.pop();
vis[x]=false;
for(i=head[x];i!=-1;i=p[i].next)
{
if(p[i].flow&&d[y=p[i].v]>d[x]+p[i].cost)
{
d[y]=d[x]+p[i].cost;
pre[y]=i;
if(vis[y])
continue;
vis[y]=true;
q.push(y);
}
}
}
return d[e]!=d[e+1];
}
int mcmf(){
int mincost=0,maxflow=0,minflow,i;
while(spfa()){
minflow=INF;
for(i =pre[e];i!=-1;i=pre[p[i].u])
minflow=min(minflow,p[i].flow);
for(i=pre[e];i!=-1;i=pre[p[i].u])
{
p[i].flow-=minflow;
p[i^1].flow+=minflow;
}
mincost+=d[e]*minflow;
maxflow+=minflow;
}
//跑完所有最短增广路
if(maxflow==2)//判读是否满流 (根据题意限制最大流为2);
return mincost;
return -1;
}
int n,m;
int main(){
cin>>n>>m;
init();//初始化
s=0;//源点为0
e=n+1;//汇点为N+1
add(s,1,2,0);//源点向点1件容量为2,花费为0的边
add(n,e,2,0);//点N向汇点建容量为2,花费为0的边
while(m--){
int u,v,c;
scanf("%d%d%d",&u,&v,&c);
add(u,v,1,c);//每条边的起点和终点建容量为1,花费为边权的的边
add(v,u,1,c);//无向边 把无向图拆成两个方向不同的有向边
}
int ans=mcmf();
printf("%d\n",ans);
return 0;
}