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  • light oj 1094 Farthest Nodes in a Tree(树的直径模板)

    1094 - Farthest Nodes in a Tree
    Time Limit: 2 second(s) Memory Limit: 32 MB

    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-1lines 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

    Output for Sample Input

    2

    4

    0 1 20

    1 2 30

    2 3 50

    5

    0 2 20

    2 1 10

    0 3 29

    0 4 50

    Case 1: 100

    Case 2: 80

    Notes

    Dataset is huge, use faster i/o methods.

    先一次搜索搜索到最远的端点。然后再从这个端点搜索整个树。即为树中最远的距离,就是树的直径

    PROBLEM SETTER: JANE ALAM JAN

    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #include<queue>
    using namespace std;
    #define M 31000
    struct node {
    	int v,next,w;
    }mp[M*3];
    int cnt,head[M],dis[M],vis[M];
    int ans,last;
    void add(int u,int v,int w){
    	mp[cnt].v=v;
    	mp[cnt].w=w;
    	mp[cnt].next=head[u];
    	head[u]=cnt++;
    }
    void bfs(int s){
    	queue<int> q;
    	memset(vis,0,sizeof(vis));
    	memset(dis,0,sizeof(dis));
    	vis[s]=1;
    	last=s;
    	ans=0;
    	q.push(s);
    	while(!q.empty()){
    		int u=q.front();
    		q.pop();
    		for(int i=head[u];i!=-1;i=mp[i].next){
    			int v=mp[i].v;
    			if(!vis[v] && dis[v]<dis[u]+mp[i].w){
    				vis[v]=1;
    				dis[v]=dis[u]+mp[i].w;
    				if(ans<dis[v]){
    					ans=dis[v];
    					last=v;
    				}
    				q.push(v);
    			}
    		}
    	}
    }
    int main(){
    	int t,n,a,b,c,k=1;
    	scanf("%d",&t);
    	while(t--){
    		scanf("%d",&n);
    		cnt=0;
    		memset(head,-1,sizeof(head));
    		for(int i=0;i<n-1;i++){
    			scanf("%d %d %d",&a,&b,&c);
    			add(a,b,c);
    			add(b,a,c);
    		}
    		bfs(0);
    		bfs(last);
    		printf("Case %d: ", k++); 
    		printf("%d
    ",ans); 
    	}
    	return 0;
    } 


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  • 原文地址:https://www.cnblogs.com/claireyuancy/p/6917327.html
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