最短路 - A - Til the Cows Come Home POJ - 2387

最新推荐文章于 2020-07-10 23:31:10 发布

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本文介绍了一道经典的图论问题——寻找两点间的最短路径。问题背景为一头名为Bessie的奶牛需要从牧场返回谷仓，通过构建图模型并使用Dijkstra算法找到了最优路径。文中提供了一份详细的AC代码实现。

A - Til the Cows Come Home

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

题意及题解：找最短路模板题，注意双向以及有多条路。

AC代码：

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn = 1005;
const int inf = 1e9;
int t,n,m;
int a[maxn][maxn];
int vis[maxn];
int length[maxn];
void intc()
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
    {
        if(i==j)a[i][j]=0;
        else a[i][j]=inf;
    }
}
void dis()
{
    for(int i=1;i<=n;i++)
    {
        length[i]=a[1][i];
        vis[i]=0;
    }
    for(int i=1;i<=n;i++)
    {
        int u=inf;
        int v;
        for(int j=1;j<=n;j++)
        {
            if(vis[j]==0&&length[j]<u)
            {
                u=length[j];
                v=j;
            }
        }
        vis[v]=1;
        for(int j=1;j<=n;j++)
        {
            length[j]=min(length[j],length[v]+a[v][j]);
        }

    }
           printf("%d\n",length[n]);
}
int main()
{
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        intc();
        for(int i=1;i<=m;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            if(a[x][y]>z)a[x][y]=a[y][x]=z;
        }
        dis();
    }
}