bzoj1752 [Usaco2005 qua]Til the Cows Come Home

Bessie返家最短路径算法

最新推荐文章于 2019-02-10 20:36:23 发布

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最新推荐文章于 2019-02-10 20:36:23 发布

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原文链接：http://www.cnblogs.com/zhber/p/4036054.html

描述了如何利用单源最短路算法帮助Bessie以最快速度返回牛棚的故事。

Description

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it. Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

Input

* Line 1: Two integers: T and N * Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

Output

* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample Input

5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100

INPUT DETAILS:

There are five landmarks.

Sample Output

90

OUTPUT DETAILS:

Bessie can get home by following trails 4, 3, 2, and 1.

这题各种单源最短路都能过

#include<cstdio>
#include<cstring>
int n,m,x,y,z,cnt,t,w=1;
int head[10001];
struct edge{
	int to,next,v;
}e[50001];
int q[50001];
bool mark[50001];
int dist[50001];
inline void ins(int u,int v,int w)
{
	e[++cnt].v=w;
	e[cnt].to=v;
	e[cnt].next=head[u];
	head[u]=cnt;
}
void insert(int u,int v,int w)
{
	ins(u,v,w);
	ins(v,u,w);
}
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;
}
inline void spfa()
{
	q[1]=1;
	mark[1]=1;
	dist[1]=0;
	while (t<w)
	{
		int now=q[++t];
		for (int i=head[now];i;i=e[i].next)
		  if (dist[now]+e[i].v<dist[e[i].to])
		  {
		  	dist[e[i].to]=dist[now]+e[i].v;
		  	if (!mark[e[i].to])q[++w]=e[i].to;
		  }
		mark[now]=0;
	}
}
int main()
{
	m=read();n=read();
	memset(dist,127/3,sizeof(dist));
	for(int i=1;i<=m;i++)
	  {
		x=read();y=read();z=read();
		insert(x,y,z);
	  }
	spfa();
	printf("%d",dist[n]);
}

转载于:https://www.cnblogs.com/zhber/p/4036054.html