POJ 2387 - Til the Cows Come Home(单源最短路)

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这是一道经典的单源最短路径问题，通过Dijkstra算法找到奶牛Bessie从田间返回谷仓的最短距离。题目设定在一个包含N个地标和T条双向牛径的农场中，Bessie需要找到从地标N回到地标1（谷仓）的最短路径。

Til the Cows Come Home
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 46795 Accepted: 15924
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.

题意：
单源最短路裸题。

AC代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int INF = 0x3f3f3f3;
const int maxn = 1005;
int dis[maxn];
bool vis[maxn];
int mp[maxn][maxn];
int n,m;
void fun()
{
    for(int i = 1;i <= n;i++)   dis[i] = mp[1][i],vis[i] = 0;
    vis[1] = 1,dis[1] = 0;
    for(int i = 1;i <= n;i++)
    {
        int minn = INF;
        int index = -1;
        for(int j = 1;j <= n;j++)
            if(!vis[j] && dis[j] < minn)    minn = dis[j],index = j;
        vis[index] = 1;
        for(int j = 1;j <= n;j++)
            if(!vis[j] && dis[j] > dis[index]+mp[index][j]) dis[j] = dis[index]+mp[index][j];
    }
    printf("%d",dis[n]);
}
int main()
{
    scanf("%d%d",&m,&n);
    for(int i = 1;i <= n;i++)
        for(int j = 1;j <= n;j++)   mp[i][j] = i==j?0:INF;
    while(m--)
    {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        mp[u][v] = min(mp[u][v],w);
        mp[v][u] = min(mp[v][u],w);
    }
    fun();
    return 0;
}