LightOJ 1257 - Farthest Nodes in a Tree (II) BFS
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 <<":\n";
for(int i = 0; i < n; i++)
cout << max(dist[i], dist2[i]) << "\n";
}
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;
}
如果有错误,请指出,谢谢