Farm Tour
题目描述
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.
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.
输入
* 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.
* Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path's length.
输出
A single line containing the length of the shortest tour.
样例输入
4 5
1 2 1
2 3 1
3 4 1
1 3 2
2 4 2
样例输出
6
http://218.5.5.242:9018/JudgeOnline/problem.php?cid=1070&pid=0
网络流裸题
将源点向点１连接一条费用０，流量２的路
将点ｎ向汇点连接一条费用０，流量２的路
后跑费用流！
但原谅我英语不好
看别人翻译
结果做成了有向图。。。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define S 0
#define T n+1
#define MAXN 100005
#define INF 200000000
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
int q[100001],top,tail;
int ans=0;
bool mark[MAXN];
int n,m,cnt=1,head[MAXN],d[MAXN],from[MAXN];
struct tu{
int from,w,to,next,c;
}bian[100001];
void ins(int f,int t,int w,int c)
{
bian[++cnt].next=head[f];
head[f]=cnt;
bian[cnt].to=t;
bian[cnt].w=w;
bian[cnt].c=c;
bian[cnt].from=f;
}
void insw(int f,int t,int w,int c)
{ins(f,t,w,c);ins(t,f,0,-c);}
bool spfa()
{
for(int i=1;i<=T;i++)d[i]=INF;
top=tail=10005;q[top]=S;mark[S]=1;d[S]=0;
while(top>=tail)
{
int u=q[tail++];
for(int i=head[u];i;i=bian[i].next)
if(bian[i].w&&d[u]+bian[i].c<d[bian[i].to])
{
d[bian[i].to]=d[u]+bian[i].c;from[bian[i].to]=i;
if(!mark[bian[i].to])
{
if(d[bian[i].to]<d[q[tail]]) q[--tail]=bian[i].to;
else q[++top]=bian[i].to; mark[bian[i].to]=1;
}
}
mark[u]=0;
}
return d[T]!=INF;
}
void mcf()
{
int minn=INF;
for(int i=from[T];i;i=from[bian[i].from])
minn=min(minn,bian[i].w);
for(int i=from[T];i;i=from[bian[i].from])
{
ans+=bian[i].c;
bian[i].w-=minn;
bian[i^1].w+=minn;
}//SPFA费用流模板
}
int main()
{
n=read();m=read();
insw(S,1,2,0);
for(int i=1;i<=m;i++)
{
int f=read(),t=read(),c=read();
insw(f,t,1,c);
insw(t,f,1,c); //要连两次
}
insw(n,T,2,0);
while(spfa())mcf();
printf("%d",ans);
return 0;
}
