POJ - 2387 Til the Cows Come Home (最短路) (day_10_J)

最新推荐文章于 2022-07-24 10:30:31 发布

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最新推荐文章于 2022-07-24 10:30:31 发布

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在一片广阔的田野中，Bessie需要找到从苹果树林回到谷仓的最短路径，以确保她能获得足够的睡眠。面对众多地标和双向牛径，如何规划路线才能使Bessie快速抵达目的地？本文通过描述一个经典的最短路径问题，展示了如何利用图论和算法解决实际问题。

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Til the Cows Come Home

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 80251 Accepted: 26484
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

题目分析：

简单的最短路。

代码实现：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<stack>
#include<cstdlib>
#include<cmath>
#include<queue>
//#include <bits/stdc++.h>
using namespace std;

const int INF = 0x3f3f3f3f;
#define pf          printf
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define ms(i,j) memset(i,j,sizeof(i))

int a[1005][1005],low[1005],p[1005];
bool newv[1005];
int main()
{
    int n,m,t,tt,ttt,v,sum;
    while(~sff(m,n))
    {
        sum=0;
        fill(a[0],a[0]+1005*1005,INF);
        ms(low,0);
        ms(p,0);
        for(int i=0; i<m; i++)
        {
            sfff(t,tt,ttt);
            if(a[t][tt]>ttt)
            {
                a[t][tt]=ttt;
                a[tt][t]=ttt;
            }
        }
        ms(newv,0);
        for(int i=1; i<=n; i++)
        {
            low[i]=a[1][i];
            if(low[i]==INF)
                p[i]=-1;
            else
                p[i]=1;
        }
        newv[1]=1;
        for(int i=2; i<=n; i++)
        {
            int mini=INF;
            v=0;
            for(int j=1; j<=n; j++)
            {
                if(!newv[j]&&low[j]<mini)
                {
                    mini=low[j];
                    v=j;
                }
            }
            if(v)
            {
                newv[v]=1;
                for(int j=1; j<=n; j++)
                {
                    if(!newv[j]&&a[j][v]<INF&&low[v]+a[v][j]<low[j])
                    {
                        low[j]=a[j][v]+low[v];
                        p[j]=v;
                    }
                }
            }
        }
        int x=n;
        while(x!=1)
        {
            sum+=a[x][p[x]];
            x=p[x];
        }
        pf("%d\n",sum);
    }
}