poj-2387 dijkstra算法

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.

大意是第一行给你两个数n, m，代表节点数和边数，后面m行，每行三个数a,b,c

代表a, b之间的一条边权重是c，求从最后一个点到第一个点的最小距离

典型的dijkstra算法求最短路问题，因为是最短路，所以第一个到最后也是这个距离，ac代码如下：

#include <iostream>
#include <cstdio>
#include <stdio.h>
#include <algorithm>
#include <cctype>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <sstream>
#include <iterator>
#include<cmath>
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
int maps[1005][1005], d[1005], vis[1005];
int t, n;
void init( int n)//初始化函数
{
    for (int i(1); i <= n; i++)
    {
        for (int j(1); j <= n; j++)
        {
            if (i == j)
                maps[i][j] = 0;
            else
                maps[i][j] = INF;
        }
        d[i] = INF;
    }
}
void dijkstra(int s)//dijkstra算法，s为起点
{
    memset(vis, 0, sizeof(vis));
    int cur = s;
    vis[s] = 1;
    d[s] = 0;
    for (int i(1); i <= n; i++)//n个点，每次选出一个，所以循环n次
    {
        for (int j(1); j <= n; j++)//更新剩下节点到初始节点的距离
        {
            if (!vis[j] && maps[cur][j] + d[cur] < d[j])
                d[j] = maps[cur][j] + d[cur];
        }
        int minn = INF;
        for (int j(1); j <= n; j++)//找出剩下点钟到初始节点距离最小的点cur
        {
            if (!vis[j] && d[j] < minn)
            {
                minn = d[j];
                cur = j;
            }
        }
        vis[cur] = 1;//标记为访问
    }
}
int main()
{
    while (~scanf("%d%d", &t, &n))
    {
        init(n);
        for (int i(0); i < t; i++)
        {
            int x, y, s;
            scanf("%d%d%d", &x, &y, &s);
            if( maps[x][y] > s && maps[y][x] > s)
                maps[x][y] = maps[y][x] = s;
        }
        dijkstra(1);//从第一个点开始找最短路
        printf("%d\n", d[n]);
    }
    return 0;
}

 

 

转载于:https://www.cnblogs.com/xujunming/p/7344462.html