[NC14550]旅行

一、题目

NC14550

二、思路

这题固定了三个点，起点、中转点、终点，很容易想到用dijkstra来写，但是直接一遍dijkstra并不能马上找到两条最长的路径，所以我们可以通过枚举每个点作为中转点时的情况，与这个点距离最远的两个点就是起点和终点，每次和当前的最长路径取最大值即可
时间复杂度: mlogn * n

三、代码

#include<bits/stdc++.h>
using namespace std;
typedef pair<int, int> PII;
const int N = 2020;
int h[N], e[N], ne[N], w[N], idx;
int dist[N];
bool st[N];
int n, m;
int max1, max2;
void add(int x, int y, int z){ 
	w[idx] = z; e[idx] = y; ne[idx] = h[x]; h[x] = idx ++;
}
void dijkstra(int u){
	memset(dist, 0x3f, sizeof dist);
	memset(st, false, sizeof st);
	dist[u] = 0;
	priority_queue<PII, vector<PII>, greater<PII> > heap;
	heap.push({0, u});
	while(heap.size()){
		PII k = heap.top();
		heap.pop();
		int ver = k.second, distance = k.first;
		if(st[ver]) continue;
		st[ver] = true;
		for(int i = h[ver]; i != -1; i = ne[i]){
			int j = e[i];
			if(dist[j] > distance + w[i]){
				dist[j] = distance + w[i];
				heap.push({dist[j], j});
				//cout << j << endl;
			}
		}
	}
	for(int k = 1; k <= n; k ++){ //判断哪两个点才是最大的两条路
		if(max1 < dist[k] && dist[k] != 0x3f3f3f3f){
			max2 = max1;
			max1 = dist[k];
			//cout << k << endl;
		}
		else if(max2 < dist[k] && dist[k] != 0x3f3f3f3f) max2 = dist[k];
	}
	//cout << endl;
}
int main(){
	int t;
	int a, b, c;
	int maxn = 0;
	cin >> t;
	while(t --){
		idx = 0;
		cin >> n >> m;	
		for(int i = 0; i <= n + 5; i ++){
			h[i] = -1;
			e[i] = 0;
			ne[i] = 0;
			w[i] = 0;
		}
		for(int i = 1; i <= m; i ++){
			scanf("%d%d%d", &a, &b, &c);
			add(a, b, c);
			add(b, a, c);
		}
		maxn = 0;
		for(int i = 1; i <= n; i ++){ //枚举每个中转点作为dijkstra的起点
			max1 = 0, max2 = 0;
			dijkstra(i);
			//cout << max1 << " " << max2 << endl;
			if(max1 != 0 && max2 != 0){
				maxn = max(maxn, max1 + max2);
			}
		}
		if(maxn == 0) cout << "-1" << endl;
		else cout << maxn << endl;
	}	
	return 0;
}