poj 2387 最短路水题

最新推荐文章于 2021-03-12 17:50:16 发布

Resurgence_zc

最新推荐文章于 2021-03-12 17:50:16 发布

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分类专栏：最短路文章标签： poj 最短路 acm

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本文介绍了一道经典的最短路径问题——POJ2387 Til the Cows Come Home，并提供了使用Dijkstra算法求解的详细过程及代码实现。问题要求找出从田野中的一个点回到谷仓的最短距离。

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poj2387- Til the Cows Come Home

题目：

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.

题意：

就是给一个图，让你求从1到n的最短路径。
两个点比较坑人：
1.没说是多组输入；
2.注意输入有重边，取最小的！！

思路：

用Dijkstra做的，至于要问Dijkstra是什么的话，看这里：
http://www.cnblogs.com/biyeymyhjob/archive/2012/07/31/2615833.html

代码：

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <limits.h>
#define inf 0x3f3f3f3f
using namespace std;
int maps[1010][1010];
bool vis[1010];
int val[1010];
int n,t;
void solve()
{
    for(int i=1;i<=n;i++)
        val[i]=maps[1][i];
    vis[1]=true;
    int T=n-1;
    while(T--)
    {
        int minn=INT_MAX;
        int index=0;
        for(int i=1;i<=n;i++)
            if(!vis[i]&&val[i]<minn)
            {
                minn=val[i];
                index=i;
            }
        vis[index]=true;
        for(int i=1;i<=n;i++)
            if(!vis[i]&&val[i]>maps[index][i]+val[index])
                val[i]=maps[index][i]+val[index];
    }
}
int main()
{
    while(scanf("%d%d",&t,&n)!=EOF)
    {
        memset(maps,inf,sizeof(maps));
        memset(vis,0,sizeof(vis));
        memset(val,0,sizeof(val));
        int x,y,z;
        for(int i=1;i<=t;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            maps[y][x]=maps[x][y]=(maps[x][y]<z?maps[x][y]:z);
        }
        solve();
        printf("%d\n",val[n]);
        return 0;
    }

}

水题两wa，23333