POJ-----2631树的直径

最新推荐文章于 2023-03-23 12:44:51 发布

原创最新推荐文章于 2023-03-23 12:44:51 发布 · 244 阅读

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本文介绍了一种算法，用于计算地图上两个最远村庄之间的道路距离。地图由不超过10,000个通过双向道路连接的村庄组成，确保任意两村庄间仅有一条不重复经过其它村庄的道路。算法采用两次广度优先搜索策略，首先从任意村庄出发找到一个最远村庄，然后以此村庄为起点再次进行搜索，以确定整个地图上的最大距离。

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Roads in the North

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2652		Accepted: 1309

Description

Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are build such that there is only one route from a village to a village that does not pass through some other village twice.
Given is an area in the far North comprising a number of villages and roads among them such that any village can be reached by road from any other village. Your job is to find the road distance between the two most remote villages in the area.

The area has up to 10,000 villages connected by road segments. The villages are numbered from 1.

Input

Input to the problem is a sequence of lines, each containing three positive integers: the number of a village, the number of a different village, and the length of the road segment connecting the villages in kilometers. All road segments are two-way.

Output

You are to output a single integer: the road distance between the two most remote villages in the area.

Sample Input

Sample Output

模板题，唯一恶心的地方是这道题居然没法输出，只能自己加个变量控制输出，提交时侯再删掉

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define maxn 10010
using namespace std;
int ans, num, nod, x, y, z;
int head[maxn], dis[maxn];
bool vis[maxn];
struct node{
	int from, to, val, next;
}edge[maxn*2];
void add(int u, int v, int w){
	edge[num].from = u;
	edge[num].to = v;
	edge[num].val = w;
	edge[num].next = head[u];
	head[u] = num++;
}
void bfs(int t){
	memset(vis, false, sizeof(vis));
	memset(dis, 0, sizeof(dis));
	queue<int > q;
	q.push(t);
	ans = dis[t] = 0;
	vis[t] = true;
	while(!q.empty()){
		int u = q.front();
		q.pop();
		for(int i = head[u]; i != -1; i = edge[i].next){
			int v = edge[i].to;
			if(!vis[v]){
				if(dis[v] < edge[i].val + dis[u]){
					dis[v] = edge[i].val + dis[u];
				}
				q.push(v);
				vis[v] = true;
				if(ans < dis[v]){
					ans = dis[v];
					nod = v;
				}
			}
		}
	}
}
int main(){
	int a, b, c;
	memset(head, -1, sizeof(head));
	num = 0;
	while(scanf("%d%d%d", &a, &b, &c) != EOF){
		add(a, b, c);
		add(b, a, c);
	}
	bfs(1);
	bfs(nod);
	printf("%d\n", ans);
	return 0;
}