Light OJ--1094（求树的直径）

最新推荐文章于 2019-07-15 16:31:00 发布

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最新推荐文章于 2019-07-15 16:31:00 发布

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分类专栏：最小、最大、次小生成树

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本文介绍了一种求解树的直径的经典算法。通过两次广度优先搜索找到树中最远的两个节点之间的距离。首先从任意节点开始寻找最远节点，再从此节点出发找下一个最远节点，从而得出树的直径。

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O - O

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1094

Description

Given a tree (a connected graph with no cycles), you have to find the farthest nodes in the tree. The edges of the tree are weighted and undirected. That means you have to find two nodes in the tree whose distance is maximum amongst all nodes.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with an integer n (2 ≤ n ≤ 30000) denoting the total number of nodes in the tree. The nodes are numbered from 0 to n-1. Each of the next n-1 lines will contain three integers u v w (0 ≤ u, v < n, u ≠ v, 1 ≤ w ≤ 10000) denoting that node u and v are connected by an edge whose weight is w. You can assume that the input will form a valid tree.

Output

For each case, print the case number and the maximum distance.

Sample Input

0 1 20

1 2 30

2 3 50

0 2 20

2 1 10

0 3 29

0 4 50

Sample Output

Case 1: 100

Case 2: 80

解体思路:是一道求树的直径的模板题.有结论,树的直径为(s,t),即树中相距最远的两结点之间地距离。从树中任意一点u出发,那么它到达的最远距离的点一定是s或t。然后从最远点出发，寻找下一个最远点，就能求出树的直径了。

代码如下:

#include<stdio.h>
#include<string.h>
#include<queue>
#define LL __int64
using namespace std;
int head[30010];
int vist[30010];
LL dfn[30010];
int node;
LL ans;
int num;
struct stu{
	int t,ne,c;
}edge[60010];
void inin(){
	num=0;
	memset(head,-1,sizeof(head));
}
void add(int a,int b,int c){
	stu E={b,head[a],c};
	edge[num]=E;
	head[a]=num++;
}
void bfs(int sx){//注意与SPFA的区别 
	ans=0;
	queue<int>q;
	memset(vist,0,sizeof(vist));
	memset(dfn,0,sizeof(dfn));
	vist[sx]=1;
	q.push(sx);
	while(!q.empty()){
		int v=q.front();
		q.pop();
		for(int i=head[v];i!=-1;i=edge[i].ne){
			int w=edge[i].t;
			if(!vist[w]&&dfn[w]<dfn[v]+edge[i].c){
				dfn[w]=dfn[v]+edge[i].c;
				if(dfn[w]>ans){
					ans=dfn[w];
					node=w;
				}
			    vist[w]=1;
			    q.push(w);
			}
		}
		
	}
}
int main(){
	int n,t,a,b,c,i,cas;
	cas=1;
    scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		inin();
		for(i=1;i<n;i++){
			scanf("%d%d%d",&a,&b,&c);
		    add(a,b,c);
		    add(b,a,c);
		}
		bfs(0);
		bfs(node);
		printf("Case %d: %lld\n",cas++,ans);
	}
	return 0;
}