LightOJ 1257

题目链接

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题意： 在一棵n节点的树上，求树上各点到树上某一点的最长距离。

思路：先进行一遍任意节点开始的BFS，找到最远的节点，在以该点为起始点BFS，得到各个点到它的距离，并找到离他最远的节点，可以证明，树上任意一节点的最远距离，要么是到第一次找的最远点的距离，要么是第二次找的最远点的距离。所以以第二次为起始点进行一遍BFS，得到各点到他的距离，就得到答案。

关闭C++流同步会出现RunTime Error

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
#include <stack>
#include <string>
#include <math.h>
#include <bitset>
#include <ctype.h>
#define IO_FOR_CPP ios_base::sync_with_stdio(false)
using namespace std;
typedef pair<int,int> P;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-9;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
int t, n, u, v, w, kase = 0;
struct Edge
{
    int u,v,w;
    Edge(){}
    Edge(int a, int b, int c):u(a),v(b),w(c){}
};
vector<int> G[N];
vector<Edge> edges;
void init(int n)
{
    for(int i = 0; i <= n; i++) G[i].clear();
    edges.clear();
}
void addEdge(int u, int v, int w)
{
    edges.push_back(Edge(u, v, w));
    edges.push_back(Edge(v, u, w));
    int m = edges.size();
    G[u].push_back(m-2);
    G[v].push_back(m-1);
}
int dist[N], dist2[N], vis[N];
void bfs(int s)
{
    memset(dist, 0, sizeof(dist));
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(s); vis[s] = 1; dist[s] = 0;

    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i = 0; i < G[u].size(); i++)
        {
            Edge &e = edges[G[u][i]];
            int v = e.v;
            if(vis[v]) continue;
            vis[v] = 1;
            dist[v] = dist[u] + e.w;
            Q.push(v);
        }
    }
}


void solve()
{
    bfs(0);
    int mmax = -INF, index = -1;
    for(int i = 0; i < n; i++)
        if(index == -1 || mmax < dist[i]) mmax = dist[index = i];
    bfs(index);
    index = -1;
    for(int i = 0; i < n; i++)
    {
        dist2[i] = dist[i];
        if(index == -1 || mmax < dist[i]) mmax = dist[index = i];
    }
    bfs(index);
    cout << "Case " << ++kase <<":
";
    for(int i = 0; i < n; i++)
        cout << max(dist[i], dist2[i]) << "
";
}

int main()
{
    //IO_FOR_CPP;
    cin >> t;
    while(t--){
        cin >> n;
        init(n);
        for(int i = 0; i < n-1; i++){
            cin >> u >> v >> w;
            addEdge(u, v, w);
        }
        solve();
    }
    return 0;
}