Description

When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000) paths that connect the fields in various ways. Each path connects two different fields and has a nonzero length smaller than 35,000.

To show off his farm in the best way, he walks a tour that starts at
his house, potentially travels through some fields, and ends at the
barn. Later, he returns (potentially through some fields) back to his
house again.

He wants his tour to be as short as possible, however he doesn't
want to walk on any given path more than once. Calculate the shortest
tour possible. FJ is sure that some tour exists for any given farm.

//最小费用流

//问题描述:有n个点m条边组成的农场,有正边权,fj闲着没事干带游客绕农场兜一圈

//他希望这次旅行越短越好,每条边部能重复走,问题等价于在网络中找两条互不相交的s-t的路线

//用费用流求解,把双向边看成两条单向的,容量为1,费用为边权,增加两个点s,t

//s到1容量为2,n到t容量为2,然后求解即可

Input

* Line 1: Two space-separated integers: N and M.

* Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path's length.

Output

A single line containing the length of the shortest tour.

Sample Input

4 5
1 2 1
2 3 1
3 4 1
1 3 2
2 4 2

Sample Output

6

Source

最小费用流

题解:就在description里。。。。

code:

spfa费用流:

 #include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#define maxn 2005
#define maxm 40050
#define INF 1061109567
using namespace std;
char ch;
int n,m,a,b,c;
struct mincost_flow{
int st,en,tot,now[maxn],son[maxm],pre[maxm],val[maxm],cost[maxm];
int dist[maxn],list[maxn],pe[maxn],head,tail;
bool bo[maxn];
int flow,sum;
void clear(){
tot=;
memset(now,,sizeof(now));
}
void put(int a,int b,int c,int d){
pre[++tot]=now[a];
now[a]=tot;
son[tot]=b;
val[tot]=c;
cost[tot]=d;
}
bool spfa(){
memset(bo,,sizeof(bo));
memset(pe,,sizeof(pe));
memset(dist,,sizeof(dist));
list[]=st,bo[st]=true,dist[st]=,head=,tail=;
while (head!=tail){
if (++head==maxn) head=;
int u=list[head];
for (int p=now[u],v=son[p];p;p=pre[p],v=son[p])
if (dist[v]>dist[u]+cost[p]&&val[p]){
dist[v]=dist[u]+cost[p],pe[v]=p;
if (!bo[v]){
if (++tail==maxn) tail=;
list[tail]=v,bo[v]=true;
}
}
bo[u]=false;
}
if (dist[en]==INF) return false;
int tmp=INF,t=;
for (int u=en,p=pe[u];u!=st;u=son[p^],p=pe[u]) tmp=min(tmp,val[p]);
for (int u=en,p=pe[u];u!=st;u=son[p^],p=pe[u]) val[p]-=tmp,val[p^]+=tmp,t+=cost[p];
flow+=tmp,sum+=t*tmp;
return true;
}
void calc(){
flow=sum=;
for (;spfa(););
}
}T;
void read(int &x){
for (ch=getchar();!isdigit(ch);ch=getchar());
for (x=;isdigit(ch);x=x*+ch-'',ch=getchar());
}
int main(){
read(n),read(m);
T.clear(),T.st=,T.en=n+,T.put(,,,),T.put(,,,),T.put(n,n+,,),T.put(n+,n,,);
for (int i=;i<=m;i++) read(a),read(b),read(c),T.put(a,b,,c),T.put(b,a,,-c),T.put(b,a,,c),T.put(a,b,,-c);
T.calc();
printf("%d\n",T.sum);
return ;
}

zkw费用流:

 #include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#define maxn 1005
#define maxm 40010
#define inf 1061109567
using namespace std;
char ch;
bool ok;
void read(int &x){
for (ok=,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=;
for (x=;isdigit(ch);x=x*+ch-'',ch=getchar());
if (ok) x=-x;
}
int n,m,a,b,d;
struct zkw_costflow{
int s,t,tot,now[maxn],son[maxm],pre[maxm],val[maxm],cost[maxm];
int dis[maxn],totflow,totcost,tmp,sla[maxn];
bool bo[maxn];
void init(){s=,t=n+,tot=,memset(now,,sizeof(now));}
void put(int a,int b,int c,int d){pre[++tot]=now[a],now[a]=tot,son[tot]=b,val[tot]=c,cost[tot]=d;}
void add(int a,int b,int c,int d){put(a,b,c,d),put(b,a,,-d);}
int dfs(int u,int rest,int totval){
if (u==t){totcost+=rest*totval;return rest;}
int ans=; bo[u]=;
for (int p=now[u],v=son[p];p&&rest;p=pre[p],v=son[p])
if (val[p]&&!bo[v]){
int d=cost[p]+dis[u]-dis[v];
if (!d){
int d=dfs(v,min(rest,val[p]),totval+cost[p]);
val[p]-=d,val[p^]+=d,ans+=d,rest-=d;
}
else sla[v]=min(sla[v],d);
}
return ans;
}
bool relax(){
int d=inf;
for (int u=s;u<=t;u++) if (!bo[u]) d=min(d,sla[u]);
if (d==inf) return false;
for (int u=s;u<=t;u++) if (!bo[u]) dis[u]+=d;
return true;
}
void work(){
memset(dis,,sizeof(dis)),totflow=totcost=;
do{
memset(sla,,sizeof(sla));
do{
memset(bo,,sizeof(bo));
tmp=dfs(s,inf,),totflow+=tmp;
}while (tmp);
}while (relax());
}
}f;
int main(){
read(n),read(m),f.init(),f.add(f.s,,,),f.add(n,f.t,,);
for (int i=;i<=m;i++) read(a),read(b),read(d),f.add(a,b,,d),f.add(b,a,,d);
f.work();
printf("%d\n",f.totcost);
return ;
}

消负圈费用流:

 #include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#define maxn 1005
#define maxm 40010
#define inf 1061109567
using namespace std;
char ch;
bool ok;
void read(int &x){
for (ok=,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=;
for (x=;isdigit(ch);x=x*+ch-'',ch=getchar());
if (ok) x=-x;
}
int n,m,a,b,d;
struct costflow{
int s,t,tot,now[maxn],son[maxm],pre[maxm],val[maxm],cost[maxm];
int dis[maxn],head,tail,list[maxn];
int top,stack[maxn],idx,path[maxn];
bool in[maxn],bo[maxn];
int totflow,totcost;
void init(){s=,t=n+,tot=,memset(now,,sizeof(now));}
void put(int a,int b,int c,int d){pre[++tot]=now[a],now[a]=tot,son[tot]=b,val[tot]=c,cost[tot]=d;}
void add(int a,int b,int c,int d){put(a,b,c,d),put(b,a,,-d);}
bool bfs(){
memset(bo,,sizeof(bo));
head=,tail=,list[]=s,dis[s]=,bo[s]=;
while (head<tail){
int u=list[++head];
for (int p=now[u],v=son[p];p;p=pre[p],v=son[p])
if (val[p]&&!bo[v]) bo[v]=,dis[v]=dis[u]+,list[++tail]=v;
}
return bo[t];
}
int dfs(int u,int rest,int totval){
if (u==t){totcost+=rest*totval;return rest;}
int ans=;
for (int p=now[u],v=son[p];p&&rest;p=pre[p],v=son[p])
if (val[p]&&dis[v]==dis[u]+){
int d=dfs(v,min(rest,val[p]),totval+cost[p]);
val[p]-=d,val[p^]+=d,ans+=d,rest-=d;
}
if (!ans) dis[u]=-;
return ans;
}
void dinic(){
totflow=totcost=;
while (bfs()) totflow+=dfs(s,inf,);
}
int spfa(int u){
in[u]=;
for (int p=now[u],v=son[p];p;p=pre[p],v=son[p])
if (val[p]&&dis[v]>dis[u]+cost[p]){
dis[v]=dis[u]+cost[p],path[v]=p;
if (in[v]) return v;
int t=spfa(v); if (t!=-) return t;
}
in[u]=; return -;
}
int find(){
memset(in,,sizeof(in));
memset(dis,,sizeof(dis));
memset(path,-,sizeof(path));
for (int i=s;i<=t;i++){
int u=spfa(i);
if (u!=-) return u;
}
return -;
}
void relax(int t){
int flow=inf,sum=,u;
for (u=t;son[path[u]^]!=t;u=son[path[u]^]) flow=min(flow,val[path[u]]),sum+=cost[path[u]];
flow=min(flow,val[path[u]]),sum+=cost[path[u]];
for (u=t;son[path[u]^]!=t;u=son[path[u]^]) val[path[u]]-=flow,val[path[u]^]+=flow;
val[path[u]]-=flow,val[path[u]^]+=flow,totcost+=flow*sum;
}
void work(){
dinic();
for (int t=find();t!=-;t=find()) relax(t);
}
}f;
int main(){
read(n),read(m),f.init(),f.add(f.s,,,),f.add(n,f.t,,);
for (int i=;i<=m;i++) read(a),read(b),read(d),f.add(a,b,,d),f.add(b,a,,d);
f.work();
printf("%d\n",f.totcost);
return ;
}

老oj2146 && Pku2135 Farm Tour的更多相关文章

  1. POJ2135 Farm Tour

      Farm Tour Time Limit: 2MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u Description ...

  2. 网络流(最小费用最大流):POJ 2135 Farm Tour

    Farm Tour Time Limit: 1000ms Memory Limit: 65536KB This problem will be judged on PKU. Original ID: ...

  3. POJ Farm Tour

    Farm Tour 题目: 约翰有N块地,家在1号,而N号是个仓库.农场内有M条道路(双向的),道路i连接这ai号地和bi号地,长度为ci. 约翰希望依照从家里出发,经过若干地后达到仓库.然后再返回家 ...

  4. [网络流]Farm Tour(费用流

    Farm Tour 题目描述 When FJ's friends visit him on the farm, he likes to show them around. His farm compr ...

  5. POJ 2135 Farm Tour (网络流,最小费用最大流)

    POJ 2135 Farm Tour (网络流,最小费用最大流) Description When FJ's friends visit him on the farm, he likes to sh ...

  6. Farm Tour(最小费用最大流模板)

    Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18150   Accepted: 7023 Descri ...

  7. POJ2135 Farm Tour —— 最小费用最大流

    题目链接:http://poj.org/problem?id=2135 Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submis ...

  8. poj 2351 Farm Tour (最小费用最大流)

    Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17230   Accepted: 6647 Descri ...

  9. poj 2135 Farm Tour 【无向图最小费用最大流】

    题目:id=2135" target="_blank">poj 2135 Farm Tour 题意:给出一个无向图,问从 1 点到 n 点然后又回到一点总共的最短路 ...

随机推荐

  1. Android开发学习之Adapter

    Adapter是指适配器的意思,在Android中,适配器扮演者重要的角色,是UI与Data实现绑定的一个桥梁.Adapter负责创建和显示每个项目的子View和提供对下层数据的访问.支持Adapte ...

  2. systrace跟踪 Android性能优化

    http://blog.csdn.net/oujunli/article/details/8138172 http://blog.csdn.net/oujunli/article/details/50 ...

  3. 廖雪锋笔记1---python变量类型

    整型:a/b a//b a%b 浮点型:.2 字符串: "" '' r"" r'' '''...''' r'''...'''' 变量值共享:写时复制 NULL型 ...

  4. block代码块介绍

    关于block的简单介绍 什么是block? Block是C语言的一个语法特性,同时也是C语言的运行时特性,它很像C中的函数指针,因为你可以像使用函数指针一样的去使用block对象:它也很像C++中的 ...

  5. tcmalloc资料

    1. 确定dylib在max os是可以成功的. http://lists.apple.com/archives/perfoptimization-dev/2008/Dec/msg00002.html ...

  6. CentOS 通过yum来升级php到php5.6,yum upgrade php 没有更新包怎么办?

    在文章中,我们将展示在centOS系统下如何将php升级到5.6,之前通过yum来安装lamp环境,直接升级的话,提示没有更新包,也就是说默认情况下php5.3.3是最新 1.查看已经安装的php版本 ...

  7. Apache MINA 框架之默认session管理类实现

    DefaultSocketSessionConfig 类 extends AbstractSocketSessionConfig extends AbstractIoSessionConfig imp ...

  8. Navicat Premium 自动备份mysql和sqlserver

    mysql篇: 1.点击计划 2.点击新建处理作业 3.选择需要备份的数据库,上级可用任务 4.点击保存按钮,输入保存文件名 5.保存后点击设置计划任务 6.计划里新建保存时间,应用后输入系统密码即可 ...

  9. 表达式:使用API创建表达式树(5)

    一.ConditionalExpression:表达式 生成如 IIF((a == b), "a和b相等", "a与b不相等") 式子. 使用: Paramet ...

  10. [转载]CentOS6.4+Mono3.0.7+Jexus5.2.5

    本文章来自互联网,但是本人已经在VM虚拟机里面测试成功,所以分享给大家 1.更新 yum -y update 2.安装Mono源码安装需要的库 yum -y install gcc gcc-c++ a ...