zoukankan      html  css  js  c++  java
  • Query on a tree II

    Query on a tree II

    You are given a tree (an undirected acyclic connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. Each edge has an integer value assigned to it, representing its length.
    We will ask you to perfrom some instructions of the following form:
    DIST a b : ask for the distance between node a and node b
    KTH a b k : ask for the k-th node on the path from node a to node b
    Example:
    N = 6
    1 2 1 // edge connects node 1 and node 2 has cost 1
    2 4 1
    2 5 2
    1 3 1
    3 6 2

    Path from node 4 to node 6 is 4 -> 2 -> 1 -> 3 -> 6
    DIST 4 6 : answer is 5 (1 + 1 + 1 + 2 = 5)
    KTH 4 6 4 : answer is 3 (the 4-th node on the path from node 4 to node 6 is 3)

    The first line of input contains an integer t, the number of test cases (t <= 25). t test cases follow.

    For each test case:
    In the first line there is an integer N (N <= 10000)
    In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 100000)
    The next lines contain instructions "DIST a b" or "KTH a b k"
    The end of each test case is signified by the string "DONE".
    There is one blank line between successive tests.

    For each "DIST" or "KTH" operation, write one integer representing its result.
    Print one blank line after each test.

    Input:

    1
    6
    1 2 1
    2 4 1
    2 5 2
    1 3 1
    3 6 2
    DIST 4 6
    KTH 4 6 4
    DONE

    Output:

    5
    3

    树上倍增即可
    同时倍增父亲和距离
    注意计算中的+1 -1
    
    ```c++
    #include <bits/stdc++.h>
    using namespace std;
    #define maxn (int)(1e5+10)
    #define LL long long
    
    inline int read(){
    	int rtn=0,f=1;char ch=getchar();
    	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    	while(isdigit(ch))rtn=(rtn<<1)+(rtn<<3)+ch-'0',ch=getchar();
    	return rtn*f;
    }
    
    struct node{
    	int a,b,nt,w;
    }e[maxn];
    
    LL w[maxn][21];
    int fa[maxn][21],dep[maxn],p[maxn],cnt;
    
    inline void add(int x,int y,int z){
    	e[++cnt].a=x;e[cnt].b=y;e[cnt].w=z;
    	e[cnt].nt=p[x];p[x]=cnt;
    }
    
    inline void dfs(int k){
    	for(int i=1;i<=20;i++)fa[k][i]=fa[fa[k][i-1]][i-1];
    	for(int i=1;i<=20;i++)w[k][i]=w[fa[k][i-1]][i-1]+w[k][i-1];
    	for(int i=p[k];i;i=e[i].nt){
    		int kk=e[i].b;
    		if(kk==fa[k][0])continue;
    		fa[kk][0]=k;dep[kk]=dep[k]+1;w[kk][0]=e[i].w;
    		dfs(kk);
    	}
    }
    
    inline int lca(int x,int y){
    	if(dep[x]<dep[y])swap(x,y);
    	for(int i=20;i>=0;i--){
    		if(dep[fa[x][i]]>=dep[y])x=fa[x][i];
    	} 
    	if(x==y)return y;
    	for(int i=20;i>=0;i--){
    		if(fa[x][i]!=fa[y][i])
    			x=fa[x][i],y=fa[y][i];
    	}return fa[x][0];
    }
    
    inline LL dis(int x,int y){
    	LL rtn=0;
    	int l=lca(x,y);
    	for(int i=20;i>=0;i--){
    		if(dep[fa[x][i]]>=dep[l])
    			rtn+=w[x][i],x=fa[x][i];
    	}
    	for(int i=20;i>=0;i--){
    		if(dep[fa[y][i]]>=dep[l])
    			rtn+=w[y][i],y=fa[y][i];
    	}return rtn;
    }
    
    inline int kth(int x,int y,int k){
    	int l=lca(x,y);
    	if(k<=dep[x]-dep[l]){
    		k-=1;
    		for(int i=20;i>=0;i--){
    			if(1<<i<=k)x=fa[x][i],k-=1<<i;
    		}
    		return x;
    	}
    	else if(k>dep[x]-dep[l]){
    		k=(dep[y]-dep[l]-(k-(dep[x]-dep[l])))+1;
    		for(int i=20;i>=0;i--){
    			if(1<<i<=k)y=fa[y][i],k-=1<<i;
    		}
    		return y;
    	}
    }
    
    int main(){
    	int T=read();
    	while(T--){
    		int n=read();cnt=0;
    		memset(p,0,sizeof(p));
    		memset(w,0,sizeof(w));
    		memset(fa,0,sizeof(fa));
    		for(int i=1;i<n;i++){
    			int a=read(),b=read(),w=read();
    			add(a,b,w);add(b,a,w);
    		}
    		dfs(1);
    		while(true){
    			char ch[10];scanf("%s",ch);
    			if(ch[1]=='O')break;
    			else if(ch[1]=='T'){
    				int x=read(),y=read(),k=read();
    				printf("%d
    ",kth(x,y,k));
    			}
    			else if(ch[1]=='I'){
    				int x=read(),y=read();
    				printf("%lld
    ",dis(x,y));
    			}
    		}
    	}
    	return 0;
    } 
    
  • 相关阅读:
    人与人之间的距离
    Web API之service worker
    css背景图等比例缩放,盒子随背景图等比例缩放
    js节流函数高级版
    vue实现百度搜索下拉提示功能
    vue实现图片点击放大
    css图片宽高相等设置
    sublime常用插件及配置
    css揭秘读书笔记
    webpack2.0简单配置教程
  • 原文地址:https://www.cnblogs.com/DexterYsw/p/7954838.html
Copyright © 2011-2022 走看看