POJ2387 Til the Cows Come Home

Til the Cows Come Home

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 39985 Accepted: 13595

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 51 2 202 3 303 4 204 5 201 5 100

Sample Output

90

题意：输入了N和T，输入的T组数据，三个数分别代表从 i 点到 j 点的距离。求出从第一点到N点的最短距离，

思路：比较裸的的Dijkstra，然后趁机模拟了一遍Dijkstra算法，

代码”：

#include<iostream>#include<cstdio>#include<cstring>#define MAXN 2003using namespace std;int T,N;int dis[MAXN][MAXN];  //储存距离的邻接矩阵int cost[MAXN];   //存放距离bool vis[MAXN];   //用于标记当前点是否遍历过void Dijkstra(int start){for(int i=1;i<=N;i++)cost[i]=dis[start][i];memset(vis,0,sizeof(vis));vis[start]=true;cost[start]=0;for(int i=1;i<=N;i++){int NM=0x7ffffff,k;  //设置NM为一个很大的值是为了松弛，for(int j=1;j<=N;j++){if(!vis[j] && NM>cost[j]){NM=cost[j];    //找出从这个点到其他邻接点的最短距离，并记下这个距离k=j;}}vis[k]=true;  //标记这个点已经遍历过for(int j=1;j<=N;j++){if(!vis[j] && cost[j]>(dis[k][j]+cost[k]) ) cost[j]=dis[k][j]+cost[k]; //这个每次更新了到目标点的距离  保证是最短距离}}}int main(){memset(dis,10,sizeof(dis));scanf("%d%d",&T,&N);for(int i=1;i<=N;i++)dis[i][i]=0;for(int i=1;i<=T;i++){int x,y,len;scanf("%d%d%d",&x,&y,&len);dis[x][y]=dis[y][x]=min(len,dis[x][y]);  //如果多次输入同一条边，取其最小值}Dijkstra(1);printf("%d\n",cost[N]);return 0;}