GDUT_寒假训练题解报告_图论专题_个人题解报告——题目：L - Til the Cows Come Home

GDUT_寒假训练题解报告_图论专题_个人题解报告——题目：L - 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.
  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.
裸的单源最短路，板子没的说，但是我被输入给坑死两次了，读题锅。
至于到底输入发生了什么不可描述的事情，看这段代码：

es[from-1][to-1]=min(value,es[from-1][to-1]);

对，他有可能两点之间有好几条路，你得把长的给去掉，我决定以后都要加这句话了，应该是这种图论题目的潜规则。
完整代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <climits>
#include <queue>
#include <stack>
#include <map>
//鬼畜头文件
using namespace std;
#define INF 0x3f3f3f3f
#define ULL unsigned long long
#define LL long long
//鬼畜define
//
int es[1010][1010];
int d[1010];
bool used[1010];
int main()
{
	int n,m;
	scanf("%d %d",&m,&n);
	fill(used,used+n,false);
	fill(d,d+n,INF);
	d[0]=0;
	memset(es,INF,sizeof(es));
	for(int time=0;time<m;time++)
	{
		int from,to,value;
		scanf("%d %d %d",&from,&to,&value);
		es[from-1][to-1]=min(value,es[from-1][to-1]);
		es[to-1][from-1]=es[from-1][to-1];
	}

	while(true)
	{
		int v=-1;
		for(int time=0;time<n;time++)
		{//取最短点
			if(!used[time]&&(v==-1||d[time]<d[v]))v=time;
		}
		if(v==-1)break;
		used[v]=true;
		for(int time=0;time<n;time++)
		{
			d[time]=min(d[time],d[v]+es[v][time]);
		}
	}
	printf("%d\n",d[n-1]);

    return 0;
}

来源：CSDN

作者：DevourPower

链接：https://blog.csdn.net/DevourPower/article/details/104010432