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HDU 5052 LCT

2024-07-30 19:36:23   219人阅读

Yaoge’s maximum profit

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 516    Accepted Submission(s): 150


Problem Description
Yaoge likes to eat chicken chops late at night. Yaoge has eaten too many chicken chops, so that Yaoge knows the pattern in the world of chicken chops. There are N cities in the world numbered from 1 to N . There are some roads between some cities, and there is one and only one simple path between each pair of cities, i.e. the cities are connected like a tree. When Yaoge moves along a path, Yaoge can choose one city to buy ONE chicken chop and sell it in a city after the city Yaoge buy it. So Yaoge can get profit if Yaoge sell the chicken chop with higher price. Yaoge is famous in the world. AFTER Yaoge has completed one travel, the price of the chicken chop in each city on that travel path will be increased by V .
 

Input
The first line contains an integer T (0 < T ≤ 10), the number of test cases you need to solve. For each test case, the first line contains an integer N (0 < N ≤ 50000), the number of cities. For each of the next N lines, the i-th line contains an integer Wi(0 < Wi ≤ 10000), the price of the chicken chop in city i. Each of the next N - 1 lines contains two integers X Y (1 ≤ X, Y ≤ N ), describing a road between city X and city Y . The next line contains an integer Q(0 ≤ Q ≤ 50000), the number of queries. Each of the next Q lines contains three integer X Y V(1 ≤ X, Y ≤ N ; 0 < V ≤ 10000), meaning that Yaoge moves along the path from city X to city Y , and the price of the chicken chop in each city on the path will be increased by V AFTER Yaoge has completed this travel.
 

Output
For each query, output the maximum profit Yaoge can get. If no positive profit can be earned, output 0 instead.
 

Sample Input
1
5
1
2
3
4
5
1 2
2 3
3 4
4 5
5
1 5 1
5 1 1
1 1 2
5 1 1
1 2 1
 

Sample Output
4
0
0
1
0
 

Source
2014 ACM/ICPC Asia Regional Shanghai Online
 


路径大值与小值差值的最大值，其中满足小值在大值前面出现，加入求u->:v的路径上答案，可以先u->make_root(），然后v->access(),然后输出答案就可以了。

代码：

‘

/* ***********************************************
Author :rabbit
Created Time :2014/11/2 19:01:37
File Name :4.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=100100;
struct Node *null;
struct Node{
	Node *ch[2],*fa;
	int Max,Min,mm,rmm,rev,add,val;
	inline void clear(int _val){
		fa=ch[0]=ch[1]=null;
		Max=Min=val=_val;
		rev=add=mm=rmm=0;
	}
	inline void push_up(){
		if(this==null)return;
		mm=0;
		mm=max(mm,ch[0]->mm);
		mm=max(mm,ch[1]->mm);
		mm=max(mm,max(val,ch[1]->Max)-ch[0]->Min);
		mm=max(mm,ch[1]->Max-min(val,ch[0]->Min));
		rmm=0;
		rmm=max(rmm,ch[0]->rmm);
		rmm=max(rmm,ch[1]->rmm);
		rmm=max(rmm,max(val,ch[0]->Max)-ch[1]->Min);
		rmm=max(rmm,ch[0]->Max-min(val,ch[1]->Min));
		Max=max(val,max(ch[0]->Max,ch[1]->Max));
		Min=min(val,min(ch[0]->Min,ch[1]->Min));
	}
	inline void setc(Node *p,int d){
		ch[d]=p;
		p->fa=this;
	}
	inline bool d(){
		return fa->ch[1]==this;
	}
	inline bool isroot(){
		return fa==null||fa->ch[0]!=this&&fa->ch[1]!=this;
	}
	inline void flip(){
		if(this==null)return;
		swap(ch[0],ch[1]);
		rev^=1;
		swap(mm,rmm);
	}
	inline void update_add(int w){
		if(this==null)return;
		Max+=w;
		Min+=w;
		val+=w;
		add+=w;
	}
	inline void push_down(){
		if(add){
			ch[0]->update_add(add);
			ch[1]->update_add(add);
			add=0;
		}
		if(rev){
			ch[0]->flip();
			ch[1]->flip();
			rev=0;
		}
	}
    inline void go(){
		if(!isroot())fa->go();
		push_down();
	}
	inline void rot(){
		Node *f=fa,*ff=fa->fa;
		int c=d(),cc=fa->d();
		f->setc(ch[!c],c);
		this->setc(f,!c);
		if(ff->ch[cc]==f)ff->setc(this,cc);
		else  this->fa=ff;
		f->push_up();
	}
	inline Node *splay(){
		go();
		while(!isroot()){
			if(!fa->isroot())
				d()==fa->d()?fa->rot():rot();
			rot();
		}
		push_up();
		return this;
	}
	inline Node *access(){
		for(Node *p=this,*q=null;p!=null;q=p,p=p->fa){
			p->splay()->setc(q,1);
			p->push_up();
		}
		return splay();
	}
	inline Node *find_root(){
		Node *x;
		for(x=access();x->push_down(),x->ch[0]!=null;x=x->ch[0]);
		return x;
	}
	void make_root(){
		access()->flip();
	}
};
Node pool[maxn],*tail;
Node*node[maxn];
struct Edge{
	int next,to;
}edge[maxn*2];
int head[maxn],tol;
inline void addedge(int u,int v){
	edge[tol].to=v;
	edge[tol].next=head[u];
	head[u]=tol++;
}
void dfs(int u,int pre){
	for(int i=head[u];i!=-1;i=edge[i].next){
		int v=edge[i].to;
		if(v==pre)continue;
		node[v]->fa=node[u];
		dfs(v,u);
	}
}
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     int n,m,T;
	 scanf("%d",&T);
	 while(T--){
		 scanf("%d",&n);
		 tail=pool;
		 null=tail++;
		 null->val=null->mm=null->rmm=0;
		 null->Max=-INF;null->Min=INF;
		 null->ch[0]=null->ch[1]=null->fa=null;
		 null->add=null->rev=0;
		 for(int i=1;i<=n;i++){
			 int w;
			 scanf("%d",&w);
			 node[i]=tail++;
			 node[i]->clear(w);
		 }
		 memset(head,-1,sizeof(head));tol=0;
		 for(int i=1;i<n;i++){
			 int u,v;
			 scanf("%d%d",&u,&v);
			 addedge(u,v);
			 addedge(v,u);
		 }
		 dfs(1,1);
		 scanf("%d",&m);
		 while(m--){
			 int u,v,d;
			 scanf("%d%d%d",&u,&v,&d);
			 node[u]->make_root();
			 node[v]->access();
			 printf("%d\n",node[v]->mm);
			 node[v]->update_add(d);
		 }
	 }
     return 0;
}


HDU 5052 LCT


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