How far away(倍增LCA)

最新推荐文章于 2020-09-23 20:48:39 发布

原创最新推荐文章于 2020-09-23 20:48:39 发布 · 232 阅读

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博客围绕HDU2586题目展开，题目描述了村庄房屋由双向道路连接，需计算两房屋间距离。给出输入输出格式及样例，题解指出是求树中两点最短距离问题，采用带权LCA，用dis数组表示节点间距离。

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传送门 HDU2586

题目描述

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this “How far is it if I want to go from house A to house B”? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path(“simple” means you can’t visit a place twice) between every two houses. Yout task is to answer all these curious people.

输入

First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

输出

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

样例

Input
2
3 2
1 2 10
3 1 15
1 2
2 3

2 2
1 2 100
1 2
2 1

Output
10
25
100
100

题解

题意：给一棵树，求两点的最短距离。
带权LCA。在原有的基础上用dis[i][j]表示节点i到i+(2^j)节点的距离，dis[i][0]就是i节点到父节点的距离，dis[i][j]=dis[i][j-1]+dis[father[i][j-1]][j-1]

Code

#include<bits/stdc++.h>
#define INIT(a,b) memset(a,b,sizeof(a))
#define LL long long
using namespace std;
const int maxn=4e4+7;
const int maxm=4e4+7;
const int inf=1<<32-1;
int Begin[maxn],fa[maxn][30],dis[maxn][30],lg[maxn],depth[maxn];
struct Edge{
	int to,val,next;
}ae[maxn<<1];
int tot=0,n,m;
void Add(int x,int y,int w){
	ae[tot]=(Edge){y,w,Begin[x]};
	Begin[x]=tot++;
}
void Dfs(int now,int f){
	depth[now]=depth[f]+1;
	fa[now][0]=f;
	for(int i=1;(1<<i)<=depth[now];i++){
		fa[now][i]=fa[fa[now][i-1]][i-1];
		dis[now][i]=dis[now][i-1]+dis[fa[now][i-1]][i-1];
	}
	for(int i=Begin[now];~i;i=ae[i].next){
		if(ae[i].to!=f){
			dis[ae[i].to][0]=ae[i].val;
			Dfs(ae[i].to,now);
		}
	}
}
int Lca(int x,int y){
	int ans=0;
	if(depth[x]<depth[y]) swap(x,y);
	while(depth[x]>depth[y]){
		int k=lg[depth[x]-depth[y]]-1;
		ans+=dis[x][k];
		x=fa[x][k];
	}
	if(x==y)return ans;
	for(int i=lg[depth[x]];i>=0;i--)
		if(fa[x][i]!=fa[y][i]){
			ans+=dis[x][i];ans+=dis[y][i];
			x=fa[x][i],y=fa[y][i];
		}
	return ans+dis[x][0]+dis[y][0];
}
int main(){
	int t;
	for(int i=1;i<maxn;i++)
		lg[i]=lg[i-1]+(1<<lg[i-1]==i);
	scanf("%d",&t);
	while(t--){
		scanf("%d %d",&n,&m);
		INIT(Begin,-1);
		int x,y,z;
		for(int i=0;i<n-1;i++){
			scanf("%d %d %d",&x,&y,&z);
			Add(x,y,z);Add(y,x,z);
		}
		Dfs(1,0);
		while(m--){
			scanf("%d %d",&x,&y);
			printf("%d\n",Lca(x,y));
		}
	}
	return 0;
}
/*
10
15 100
1 2 2
1 3 3
2 4 4
2 5 5
2 6 6
3 7 7
3 8 8
8 9 9
8 10 10
8 11 11
10 12 12
10 13 13
12 14 14
12 15 15
4 15
*/