Farthest Nodes in a Tree

最新推荐文章于 2019-06-28 18:16:29 发布

原创最新推荐文章于 2019-06-28 18:16:29 发布 · 269 阅读

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本文介绍了一道算法题目：在一个带权无向树中找到距离最远的两个节点，并给出了详细的输入输出样例及C++实现代码。

Farthest Nodes in a Tree

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

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

怎么说呢，注意数组，别再开小了。。不然又是Runtime Error！！大致上跟Cow Marathon题一样。

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define MAXN 60000+10
using namespace std;
struct Edge {
    int from, to, val, next;
}edge[MAXN * 2];
int head[MAXN];
int edgenum;
void init(){
	memset(head, -1, sizeof(head)); 
	edgenum = 0; 
}
void addEdge(int u, int v, int w) {
    Edge E = {u,v,w,head[u]};
    edge[edgenum]=E;
    head[u] = edgenum++;
}
int ans;
int Tnode;
int dist[MAXN];
bool vis[MAXN];
int n;
void BFS(int s) {
    memset(dist, 0, sizeof(dist)); 
	memset(vis, false, sizeof(vis));
    queue<int> Q; 
	Q.push(s); 
	vis[s] = true; 
	dist[s] = 0; ans = 0;
    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]&&dist[v] < dist[u] + edge[i].val) {
                vis[v] = true;
                dist[v] = dist[u] + edge[i].val;
                if(dist[v] > ans) {
            		ans = dist[v];
            		Tnode = v;
        		}
                Q.push(v);
            }
        }
    }
}
int main(){
	int t,a,b,c,count=1;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		init();
		for(int i=1;i<n;i++){
			scanf("%d%d%d",&a,&b,&c);
			a++;b++;
			addEdge(a,b,c);
			addEdge(b,a,c);
		}
		BFS(1);
		BFS(Tnode);
		printf("Case %d: %d\n",count++,ans);
	}
}