ZOJ 3195 Design the city（LCA变形）

标签：zoj   lca   

题意：给定一棵树，求连接三点所需的最短距离。

思路：LCA变形，连接三点的最短距离可以转化为求任意两点距离之和的和再除以二。

#include<cstdio>  
#include<cstring>  
#include<cmath>  
#include<cstdlib>  
#include<iostream>  
#include<algorithm>  
#include<vector>  
#include<map>  
#include<queue>  
#include<stack> 
#include<string>
#include<map> 
#include<set>
#define eps 1e-6
#define pii (pair<int, int>) 
#define LL long long  
using namespace std;  
//#pragma comment(linker, "/STACK:1024000000,1024000000") 
const int maxn = 60000;
const int maxq = 75000;
//const int INF = 0x3f3f3f3f;
int dist[maxn], pnt[maxn];
int ans[3*maxq];
bool vis[maxn];
vector<int> G[maxn], w[maxn], num[maxn], query[maxn];
int n, m;
int find(int x) {
	if(x == pnt[x]) return x;
	return pnt[x] = find(pnt[x]);
}
void dfs(int u, int val) {
	dist[u] = val; vis[u] = 1; pnt[u] = u;
	int sz1 = G[u].size();
	for(int i = 0; i < sz1; i++) {
		int v = G[u][i];
		if(vis[v]) continue;
		dfs(v, val+w[u][i]);
		pnt[v] = u;
	}
	int sz2 = query[u].size();
	for(int i = 0; i < sz2; i++) {
		int v = query[u][i];
		if(vis[v]) ans[num[u][i]] = dist[u]+dist[v]-2*dist[find(v)];
	}
} 
void init() {
	memset(vis, 0, sizeof(vis));
	for(int i = 0; i < n; i++) {
		G[i].clear(); 
		w[i].clear(); 
		query[i].clear(); 
		num[i].clear();
	}
}
int kase;
int main() {
//	freopen("input.txt", "r", stdin);
	while(scanf("%d", &n) == 1) {
		if(kase++) puts("");
		init();
		for(int i = 0; i < n-1; i++) {
			int u, v, d;
			scanf("%d%d%d", &u, &v, &d);
			G[u].push_back(v);
			G[v].push_back(u);
			w[u].push_back(d);
			w[v].push_back(d);
		}
		cin >> m;
		for(int i = 0; i < m; i++) {
			int u, v, t;
			scanf("%d%d%d", &u, &v, &t);
			query[u].push_back(v); query[u].push_back(t);
			query[v].push_back(u); query[v].push_back(t);
			query[t].push_back(u); query[t].push_back(v);
			num[u].push_back(3*i); num[u].push_back(3*i+1);
			num[v].push_back(3*i); num[v].push_back(3*i+2);
			num[t].push_back(3*i+1); num[t].push_back(3*i+2);
		}
		dfs(0, 0);
		for(int i = 0; i < m; i++) printf("%d\n", (ans[3*i]+ans[3*i+1]+ans[3*i+2])/2);
	}
	return 0;
}

ZOJ 3195 Design the city（LCA变形）

标签：zoj   lca   

原文地址：http://blog.csdn.net/u014664226/article/details/47261603