P1119 灾后重建

链接

https://www.luogu.com.cn/problem/P1119

题目

知识点

floyd算法

思路

看题解,讲的差不多,本篇就是记录下写过的题。 唯一要注意的就是当遍历k(本代码用cnt代替)时,ij都要从0取到n-1。

代码

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<string.h>
#include<limits.h>
#include<string>
#include<vector>
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
#define int long long 
const int INF = 1e9;
const int N = 210;
int times[N];
int dp[N][N];


signed main()
{
	IOS;
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < N; i++)
		for (int j = 0; j < N; j++)
			dp[i][j] = INF;
	for (int i = 0; i < n; i++)cin >> times[i];
	for (int i = 0; i < m; i++)
	{
		int a, b, v;
		cin >> a >> b >> v;
		dp[a][b] = dp[b][a] = v;
	}
	int q; cin >> q;
	int cnt = 0;
	while (q--)
	{
		int x, y, t; cin >> x >> y >> t;
		while (cnt<n and times[cnt] <= t)
		{
            //这里很关键
			for (int i = 0; i < n; i++)
				for (int j = 0; j < n; j++)
					dp[i][j] = min(dp[i][j], dp[i][cnt] + dp[cnt][j]);
			cnt++;
		}
		if (dp[x][y] == INF or times[x]>t or times[y]>t)cout << -1;
		else cout << dp[x][y];
		cout << '\n';
	}

	return 0;
}