POJ 2631 -- Roads in the North【树的直径 && 裸题】

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Description

Input

Output

Sample Input

5 1 6
1 4 5
6 3 9
2 6 8
6 1 7

Sample Output

22

裸树的直径有木有！！！

#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#define maxn 40000+10
#define maxm 80000+10
using namespace std;
int ed, sum;
int dist[maxn], vis[maxn];
int head[maxn], cnt;

struct node {
    int u, v, w, next;
};
node edge[maxm];

void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int w){
    edge[cnt] = {u, v, w, head[u]};
    head[u] = cnt++;
    edge[cnt] = {v, u, w, head[v]};
    head[v] = cnt++;
}

void bfs(int sx){
    queue<int>q;
    memset(vis, 0, sizeof(vis));
    memset(dist, 0, sizeof(dist));
    ed = sx;
    sum = 0;
    vis[sx] = 1;
    q.push(sx);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        for(int i = head[u]; i != -1; i = edge[i].next){
            int v = edge[i].v;
            if(!vis[v]){
                dist[v] = dist[u] + edge[i].w;
                if(sum < dist[v]){
                    ed = v;
                    sum = dist[v];
                }
                vis[v] = 1;
                q.push(v);
            }
        }
    }
}

int main (){
    int a, b, c;
    init();
    while(scanf("%d%d%d", &a, &b, &c) != EOF)
        add(a, b, c);
    bfs(1);
    bfs(ed);
    printf("%d\n", sum);
    return 0;
}

POJ 2631 -- Roads in the North【树的直径 && 裸题】

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原文地址：http://blog.csdn.net/hpuhjh/article/details/47728449