SPOJ Problem Set (classical)

375. Query on a tree

Problem code: QTREE

You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.

We will ask you to perfrom some instructions of the following form:

• CHANGE i ti : change the cost of the i-th edge to ti
  or
• QUERY a b : ask for the maximum edge cost on the path from node a to node b

Input

The first line of input contains an integer t, the number of test cases (t <= 20). 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 <= 1000000),
• The next lines contain instructions "CHANGE i ti" or "QUERY a b",
• The end of each test case is signified by the string "DONE".

There is one blank line between successive tests.

Output

For each "QUERY" operation, write one integer representing its result.

Example

Input:
1

3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE

Output:
1
3

边权，单点修改，区间最值。

代码：

/* ***********************************************
Author :xianxingwuguan
Created Time :2014/3/1 10:06:18
File Name :1.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=60010;
struct Edge{
	int next,to;
}edge[2*maxn];
int head[maxn],tot;
int top[maxn];
int fa[maxn];
int deep[maxn];
int num[maxn];
int p[maxn],fp[maxn],son[maxn],pos;
void init(){
	tot=0;
	memset(head,-1,sizeof(head));
	pos=0;
	memset(son,-1,sizeof(son));
}
void addedge(int u,int v){
	edge[tot].to=v;
	edge[tot].next=head[u];
	head[u]=tot++;
}
void dfs1(int u,int pre,int d){
	deep[u]=d;
	fa[u]=pre;
	num[u]=1;
	for(int i=head[u];i!=-1;i=edge[i].next){
		int v=edge[i].to;
		if(v==pre)continue;
		dfs1(v,u,d+1);
		num[u]+=num[v];
		if(son[u]==-1||num[v]>num[son[u]])
			son[u]=v;
	}
}
void getpos(int u,int sp){
	top[u]=sp;
	p[u]=pos++;
	fp[p[u]]=u;
	if(son[u]==-1)return;
	getpos(son[u],sp);
	for(int i=head[u];i!=-1;i=edge[i].next){
		int v=edge[i].to;
		if(v!=son[u]&&v!=fa[u])
			getpos(v,v);
	}
}
struct Node{
	int l,r,sum;
}tree[7*maxn];
void build(int i,int l,int r){
	tree[i].l=l;
	tree[i].r=r;
	tree[i].sum=0;
	if(l==r)return;
	int mid=(l+r)/2;
	build(2*i,l,mid);
	build(2*i+1,mid+1,r);
}
void pushup(int i){
	tree[i].sum=max(tree[2*i].sum,tree[2*i+1].sum);
}
void update(int i,int k,int val){
	if(tree[i].l==tree[i].r){
		tree[i].sum=val;
		return;
	}
	int mid=(tree[i].l+tree[i].r)>>1;
	if(k<=mid)update(2*i,k,val);
	else update(2*i+1,k,val);
	pushup(i);
}
int query(int i,int l,int r){
	if(tree[i].l>=l&&tree[i].r<=r)return tree[i].sum;
	int mid=(tree[i].l+tree[i].r)/2;
	int ans=0;
	if(l<=mid)ans=max(ans,query(2*i,l,r));
	if(r>mid)ans=max(ans,query(2*i+1,l,r));
	return ans;
}
int find(int u,int v){
	int f1=top[u],f2=top[v];
	int tmp=0;
	while(f1!=f2){
		if(deep[f1]<deep[f2])
			swap(f1,f2),swap(u,v);
		tmp=max(tmp,query(1,p[f1],p[u]));
		u=fa[f1];f1=top[u];
	}
	if(u==v)return tmp;
	if(deep[u]>deep[v])swap(u,v);
	return max(tmp,query(1,p[son[u]],p[v]));
}
int e[maxn][3];
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
 	 int n,m,T;
	 scanf("%d",&T);
	 while(T--){
		 init();
		 scanf("%d",&n);
		 for(int i=0;i<n-1;i++){
			 scanf("%d%d%d",&e[i][0],&e[i][1],&e[i][2]);
			 addedge(e[i][0],e[i][1]);
			 addedge(e[i][1],e[i][0]);
		 }
		 dfs1(1,0,0);
		 getpos(1,1);
		// cout<<"111111"<<endl;
		 build(1,0,pos-1);
		// cout<<"222222"<<endl;
		 for(int i=0;i<n-1;i++){
			 if(deep[e[i][0]]>deep[e[i][1]])
				 swap(e[i][0],e[i][1]);
				 update(1,p[e[i][1]],e[i][2]);
		 }
		 char op[33];
		 while(~scanf("%s",op)){
			 int a,b;
			 if(op[0]==‘D‘)break;
			 scanf("%d%d",&a,&b);
			 if(op[0]==‘C‘)update(1,p[e[a-1][1]],b);
			 else printf("%d\n",find(a,b));
		 }
	 }
     return 0;
}

SPOJ 375 树链剖分,布布扣,bubuko.com

SPOJ 375 树链剖分

原文：http://blog.csdn.net/xianxingwuguan1/article/details/20211609