POJ 1986 Distance Queries LCA树上两点的距离

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题目来源：POJ 1986 Distance Queries

题意：给你一颗树 q次询问 每次询问你两点之间的距离

思路：对于2点 u v dis(u,v) = dis(root,u) + dis(root,v) - 2*dis(roor,LCA(u,v)) 求最近公共祖先和dis数组

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn = 40010;
int first[maxn], head[maxn], cnt, sum;
struct edge
{
	int u, v, w, next;
}e[maxn*2], qe[maxn], Q[maxn];
int ans[maxn];
int f[maxn], vis[maxn];
int d[maxn];
void AddEdge(int u, int v, int w)
{
	e[cnt].u = u;
	e[cnt].v = v;
	e[cnt].w = w;
	e[cnt].next = first[u];
	first[u] = cnt++;
	e[cnt].u = v;
	e[cnt].v = u;
	e[cnt].w = w;
	e[cnt].next = first[v];
	first[v] = cnt++;
}

void AddEdge2(int u, int v, int w)
{
	qe[sum].u = u;
	qe[sum].v = v;
	qe[sum].w = w;
	qe[sum].next = head[u];
	head[u] = sum++;
	qe[sum].u = v;
	qe[sum].v = u;
	qe[sum].w = w;
	qe[sum].next = head[v];
	head[v] = sum++;
}

int find(int x)
{
	if(f[x] != x)
		return f[x] = find(f[x]);
	return f[x];
}
void LCA(int u, int k)
{
	f[u] = u;
	d[u] = k;
	vis[u] = true;
	for(int i = first[u]; i != -1; i = e[i].next)
	{
		int v = e[i].v;
		if(vis[v])
			continue;	
		LCA(v, k + e[i].w);
		f[v] = u;
	}
	for(int i = head[u]; i != -1; i = qe[i].next)
	{
		int v = qe[i].v;
		if(vis[v])
		{
			ans[qe[i].w] = find(v);
		}
	}
} 
int main()
{
	int n, m;
	memset(first, -1, sizeof(first));
	memset(head, -1, sizeof(head));
	cnt = 0;
	sum = 0;
	scanf("%d %d", &n, &m);
	for(int i = 0; i < m; i++)
	{
		int u, v, w;
		char s[10];
		scanf("%d %d %d %s", &u, &v, &w, s);
		AddEdge(u, v, w);
	}
	int q;
	scanf("%d", &q);
	for(int i = 0; i < q; i++)
	{
		int u, v; 
		scanf("%d %d", &u, &v);
		Q[i].u = u, Q[i].v = v;
		AddEdge2(u, v, i);
		AddEdge2(v, u, i);
		
	}
	memset(vis, 0, sizeof(vis));
	d[1] = 0;
	LCA(1, 0);
	for(int i = 0; i < q; i++)
	{
		int u = Q[i].u, v = Q[i].v;
		printf("%d\n", d[u] + d[v] - 2*d[ans[i]]);
	}
	return 0;
}

POJ 1986 Distance Queries LCA树上两点的距离,布布扣,bubuko.com

POJ 1986 Distance Queries LCA树上两点的距离

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原文地址：http://blog.csdn.net/u011686226/article/details/37649873