POJ 2387 链式前向星下的SPFA

(POJ)[http://poj.org/problem?id=2387] Til the Cows Come Home Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 69789 Accepted: 23386 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
Sample Output

90
Hint

INPUT DETAILS:

There are five landmarks.

OUTPUT DETAILS:

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

USACO 2004 November

#include <stdio.h>
#include <string.h>
#include <string>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>

using namespace std;

const int maxn = 1e5;
const int maxm = 1050;
const int inf = 0x3f3f3f3f;
int n,m;

struct node
{
    int to,w,next;
}e[maxn];
int k;
int head[maxn];
int dis[maxn],vis[maxn];
void add(int u,int v,int w)
{
    e[k].to=v;
    e[k].next=head[u];
    e[k].w=w;
    head[u]=k++;
}
queue<int> q;
void spfa(int st)
{
    for(int i=1;i<=n;i++) dis[i]=inf;
    dis[st]=0;
    vis[st]=1;
    q.push(st);
    while(!q.empty())
    {
        int now = q.front();
        q.pop();
        vis[now]=0;
        for(int i=head[now]; i; i=e[i].next)
        {
            int ed = e[i].to;
            if(dis[ed] > dis[now] + e[i].w)
            {
                dis[ed] = dis[now]+e[i].w;
                if(!vis[ed])
                {
                    q.push(ed);
                    vis[ed]=1;
                }
            }
        }
    }
}
//79MS
int main()
{
    int x,y,z;
    while(cin>>m>>n)
    {
        memset(dis,inf,sizeof(dis));
        memset(vis,0,sizeof(vis));
        memset(head,-1,sizeof(head));
        k=1;
        for(int i=1;i<=m;i++)
        {
            cin>>x>>y>>z;
            add(x,y,z);
            add(y,x,z);
        }
        spfa(1);
        printf("%d
",dis[n]);
    }

}
/*
给了一个无向图，输入t和n，
t代表几个顶点，n代表问的是从第一个顶点到第n的顶点的最短距离，
各种最短路算法的模板题
Sample Input
5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100
Sample Output
90
*/