HDOJ 5372 Segment Game 树状数组+离散化

因为这题的线段长度是递增的....所以:

题解：对于新插入的线段，查询有多少个线段左端点大于等于该线段的左端点。 再查询有多少个线段的右端点大于该线段右端点， 两者之差就是答案。用两个树状数组搞定。时间复杂度nlog

Segment Game

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 975    Accepted Submission(s): 275

Problem Description

Lillian is a clever girl so that she has lots of fans and often receives gifts from her fans.

One day Lillian gets some segments from her fans Lawson with lengths of 1,2,3... and she intends to display them by adding them to a number line.At the i-th add operation,she will put the segment with length of i on the number line.Every time she put the segment on the line,she will count how many entire segments on that segment.During the operation ,she may delete some segments on the line.(Segments are mutually independent)

 

Input

There are multiple test cases.

The first line of each case contains a integer n — the number of operations(1<=n<= 2∗105 , ∑n <= 7∗105 )

Next n lines contain the descriptions of the operatons,one operation per line.Each operation contains two integers a , b. 

if a is 0,it means add operation that Lilian put a segment on the position b(|b|< 109 ) of the line.
(For the i-th add operation,she will put the segment on [b,b+i] of the line, with length of i.)

if a is 1,it means delete operation that Lilian will delete the segment which was added at the b-th add operation.

 

Output

For i-th case,the first line output the test case number.

Then for each add operation,ouput how many entire segments on the segment which Lillian newly adds.

 

Sample Input

   
   
   
   

    
    
    
    3
0 0
0 3
0 1
5
0 1
0 0
1 1
0 1
0 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1:
0
0
0
Case #2:
0
1
0
2

    
    
    
    
 
     
 
      Hint 
     
 For the second case in the sample: At the first add operation,Lillian adds a segment [1,2] on the line. At the second add operation,Lillian adds a segment [0,2] on the line. At the delete operation,Lillian deletes a segment which added at the first add operation. At the third add operation,Lillian adds a segment [1,4] on the line. At the fourth add operation,Lillian adds a segment [0,4] on the line 
    
    
    
    
     
   
   
   
   

 

Source

2015 Multi-University Training Contest 7

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月11日 星期二 20时49分58秒
File Name     :HDOJ5372.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int maxn=200200;

int n,nx;

struct CZ
{
	CZ(){}
	CZ(int _id,int _kind,int _left,int _right,int _del)
	{ 
		id=_id,kind=_kind,left=_left,right=_right,del=_del;
	}
	int id,kind,left,right,del;
}cz[maxn];

int caru_left[maxn],caru_right[maxn],cr;

int allnum[maxn*3],an;

/***********************BIT******************************/

int tree_left[maxn*3],tree_right[maxn*3];

inline int lowbit(int x) { return x&(-x); }

void add(int* tree,int p,int v)
{
	for(int i=p;i;i-=lowbit(i))
	{
		tree[i]+=v;
	}
}

int sum(int* tree,int p)
{
	int ret=0;
	for(int i=p;i<an+100;i+=lowbit(i))
	{
		ret+=tree[i];
	}
	return ret;
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	int cas=1;
	while(scanf("%d",&n)!=EOF)
	{
		an=0; nx=1; cr=1;
		for(int i=0,kind,val;i<n;i++)
		{
			scanf("%d%d",&kind,&val);
			if(kind==0)
			{
				int v1=val;
				int v2=val+nx;
				nx++;
				allnum[an++]=v1; allnum[an++]=v2;
				cz[i]=CZ(i,0,v1,v2,0);

				caru_left[cr]=v1; caru_right[cr]=v2; cr++;
			}
			else if(kind==1)
			{
				cz[i]=CZ(i,1,-999,-999,val);
			}
		}

		sort(allnum,allnum+an);
		an=unique(allnum,allnum+an)-allnum;

		//for(int i=0;i<an;i++) cout<<allnum[i]<<","; putchar(10); 

		memset(tree_left,0,sizeof(tree_left));
		memset(tree_right,0,sizeof(tree_right));

		printf("Case #%d:\n",cas++);
		for(int i=0;i<n;i++)
		{
			CZ cc=cz[i];
			if(cc.kind==0)
			{
				int l=lower_bound(allnum,allnum+an,cc.left)-allnum+1;
				int r=lower_bound(allnum,allnum+an,cc.right)-allnum+1;

				int s1=sum(tree_left,l);
				int s2=sum(tree_right,r+1);

				printf("%d\n",s1-s2);

				add(tree_left,l,1);
				add(tree_right,r,1);
			}
			else if(cc.kind==1)
			{
				cc.left=caru_left[cc.del];
				cc.right=caru_right[cc.del];

				int l=lower_bound(allnum,allnum+an,cc.left)-allnum+1;
				int r=lower_bound(allnum,allnum+an,cc.right)-allnum+1;

				add(tree_left,l,-1);
				add(tree_right,r,-1);
			}
		}
	}
    
    return 0;
}