Til the Cows Come Home POJ - 2387

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.
问题概述
第一行输入T，N接下来的T行输入点v，u以及边长v，u的权重。N代表有1~N个点。
求母牛从N走到1的最短路径。
问题分析
采用Dijstra算法，注意输入的数据中可能会有相同的边，要取权重最小的一条。
Dijstra算法描述。
集合S起始值为始发顶点N（作为新加入的顶点）
集合u起始值为除N以外的其他顶点。

1.遍历新加入到S中节点的孩子节点（只在u不在s中）跟新它们到起始顶点N的距离（一开始为正无穷)
2.将u中到达N距离最短的顶点加入到s中
3.重复步骤1与2直到u为空。
注：s中顶点存储的值为该顶点到达N的最小值
知识点介绍

truct cmp{    bool operator()(int x,int y){        return result[x]>result[y];    }};priority_queue<int,vector<int>,cmp> que;

自定义优先队列<比较。
代码实现

#include<iostream>#include<cstdio>#include<queue>#include<vector>using namespace std;#define INF 1<<29#define maxi 1002int map[maxi][maxi];int result[maxi];int s[maxi];vector<int>vec[maxi];struct cmp{    bool operator()(int x,int y){        return result[x]>result[y];    }};priority_queue<int,vector<int>,cmp> que;int dfs(int N){    int vis[maxi]={0};    vis[N]=1;    while(!que.empty()){        int temp=que.top();        que.pop();        if(temp==1)return result[temp];        s[temp]=1;        for(int i=1;i<N+1;i++)            if(map[temp][i]!=INF&&s[i]==0){                if(map[temp][i]+result[temp]<result[i])                    result[i]=map[temp][i]+result[temp];                if(vis[i]==0)que.push(i);                vis[i]=1;            }    }}int main(){    int T,N;    scanf("%d %d",&T,&N);    for(int i=1;i<N+1;i++){        result[i]=INF;        s[i]=0;        for(int j=1;j<N+1;j++)             map[i][j]=INF;    }    for(int i=1;i<T+1;i++){        int v1,v2,w;        scanf("%d %d %d",&v1,&v2,&w);        if(map[v1][v2]>w)map[v2][v1]=map[v1][v2]=w;    }    que.push(N);    result[N]=0;    printf("%d\n",dfs(N));    return 0;}