hdu 1853 最小费用流好题 环的问题
Cyclic Tour
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 1120 Accepted Submission(s): 579
Problem Description
There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom wants the total length of all the tours minimum, but he is too lazy to calculate. Can you help him?
Input
There are several test cases in the input. You should process to the end of file (EOF).
The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B, whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000).
The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B, whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000).
Output
Output one number for each test case, indicating the minimum length of all the tours. If there are no such tours, output -1.
Sample Input
6 9 1 2 5 2 3 5 3 1 10 3 4 12 4 1 8 4 6 11 5 4 7 5 6 9 6 5 4 6 5 1 2 1 2 3 1 3 4 1 4 5 1 5 6 1
Sample Output
42 -1
Hint
In the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42.
Author
RoBa@TJU
Source
题意：
有n个点 m条边
之后让这n个点组成多个环 使得每个点只能在一个环中 每个环最少2个点
求满足上面时 所有环的周长的和最小是多少
思路：
如果几个点构成一个环的话，那么这每一个点的入度与出度都是为1的 根据此 构造网络流图
设一个源点0，汇点2*n+1，源点连接每一个u,容量为1，费用为0；汇点连接每一个v+n，容量也为1，费用为0；从u到v建一条边，容量为1，费用为w；那么这就转换成了最小费用最大流的模板题，假设最后最大流为n，那么说明恰好每一个点都是入度出度为1，即构成了环。
#include <stdio.h>
#include <iostream>
#include <string.h>
#include<cmath>
using namespace std;
const int N=300;
const int MAXE=200000;
const int inf=1<<30;
int head[N],ep;
int d[N],pre[N];
bool vis[N];
int q[MAXE];
struct Edge
{
int u,v,c,w,next;
}edge[MAXE];
void addedge(int u,int v,int w,int c)//u v 费用 容量
{
edge[ep].u=u;
edge[ep].v=v;
edge[ep].w=w;
edge[ep].c=c;
edge[ep].next=head[u];
head[u]=ep++;
edge[ep].v=u;
edge[ep].u=v;
edge[ep].w=-w;
edge[ep].c=0;
edge[ep].next=head[v];
head[v]=ep++;
}
int SPFA(int src,int des)
{
int l,r;
memset(pre,-1,sizeof(pre));
memset(vis,0,sizeof(vis));
for(int i=0;i<=des;i++) d[i]=inf;
d[src]=0;
l=0;r=0;
q[r++]=src;
vis[src]=1;
while(l<r)
{
int u=q[l++];
vis[u]=0;
for(int j=head[u];j!=-1;j=edge[j].next)
{
int v=edge[j].v;
if(edge[j].c>0&&d[u]+edge[j].w<d[v])
{
d[v]=d[u]+edge[j].w;
pre[v]=j;
if(!vis[v])
{
vis[v]=1;
q[r++]=v;
}
}
}
}
if(d[des]==inf)
return 0;
return 1;
}
int flow;
int MCMF(int src,int des)
{
flow=0;int ans=0;
while(SPFA(src,des))
{
ans+=d[des];
int u=des;
int mini=inf;
while(u!=src)
{
if(edge[pre[u]].c<mini)
mini=edge[pre[u]].c;
u=edge[pre[u]].u;
}
flow+=mini;
u=des;
while(u!=src)
{
edge[pre[u]].c-=mini;
edge[pre[u]^1].c+=mini;
u=edge[pre[u]].u;
}
}
return ans;
}
int main()
{
int n,m,i,src,des;
while(scanf("%d%d",&n,&m)!=EOF)
{
ep=0;
memset(head,-1,sizeof(head));
src=0;
des=2*n+1;
while(m--)
{
int v1,v2,w;
scanf("%d %d %d",&v1,&v2,&w);
addedge(v1,v2+n,w,1);
}
for(i=1;i<=n;i++)
{
// addedge(i,i+n,0,1);
addedge(src,i,0,1);
addedge(i+n,des,0,1);
}
int ans=MCMF(src,des);
if(flow==n)
printf("%d\n",ans);
else printf("-1\n");
}
return 0;
}