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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

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

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
 

/* ***********************************************
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;
}

HDOJ 5372 Segment Game 树状数组+离散化

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原文地址：http://blog.csdn.net/ck_boss/article/details/47452501