poj2528 Mayor's posters

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 47849		Accepted: 13894

Description

The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for placing the posters and introduce the following rules: 

• Every candidate can place exactly one poster on the wall. 
  
• All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown). 
  
• The wall is divided into segments and the width of each segment is one byte. 
  
• Each poster must completely cover a contiguous number of wall segments.

They have built a wall 10000000 bytes long (such that there is enough place for all candidates). When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. Moreover, the candidates started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections. 
Your task is to find the number of visible posters when all the posters are placed given the information about posters' size, their place and order of placement on the electoral wall. 

Input

The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines describe the posters in the order in which they were placed. The i-th line among the n lines contains two integer numbers l_i and ri which are the number of the wall segment occupied by the left end and the right end of the i-th poster, respectively. We know that for each 1 <= i <= n, 1 <= l_i <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered l_i, l_i+1 ,... , ri.

Output

For each input data set print the number of visible posters after all the posters are placed. 

The picture below illustrates the case of the sample input. 

Sample Input

1
5
1 4
2 6
8 10
3 4
7 10

Sample Output

4

这道题可以用线段树做，记录每一段的涂色情况，没有涂色则为0，多种颜色则为-1，一种颜色则为value。这里离散化的时候要注意如果排序后两个数大于1，那么要在中间加一个数，如数据3 3,7 3，4 6,7，那么离散化的时候要1,2,3,4,5.不然颜色只有2中，中间的4-6会被覆盖。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
int pos[80005],a[10006],c[10005],m,sum[80006],s[80006],k;
struct node{
	int l,r,ans;
}b[16*10005];

void build(int l,int r,int i)
{
	int mid;
	b[i].l=l;b[i].r=r;b[i].ans=0;
	if(l==r)return;
	mid=(l+r)/2;
	build(l,mid,i*2);
	build(mid+1,r,i*2+1);
}

void update(int l,int r,int value,int i)
{
	int mid;
	if(b[i].l==l && b[i].r==r){
		b[i].ans=value;return;
	}
	if(b[i].ans>0){
		b[i*2].ans=b[i*2+1].ans=b[i].ans;
	}
	b[i].ans=-1;
	mid=(b[i].l+b[i].r)/2;
	if(r<=mid)update(l,r,value,i*2);
	else if(l>mid)update(l,r,value,i*2+1);
	else {
		update(l,mid,value,i*2);
		update(mid+1,r,value,i*2+1);
	}
}

void question(int i)
{
	int  mid;
	if(b[i].ans>0){
		sum[++k]=b[i].ans;return;
	}
	if(b[i].ans==0 || (b[i].l==b[i].r))return;
	else {
		question(i*2);question(i*2+1);
	}
}

int find(int x)
{
	int l=1,r=m,mid;
	while(l<=r){
		mid=(l+r)/2;
		if(pos[mid]>x)r=mid-1;
		else if(pos[mid]<x)l=mid+1;
		else return mid;
	}
	if(pos[l]==x)return l;
	else return r;
}

int main()
{
	int n,i,j,T,l,r,num;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		for(i=1;i<=n;i++){
			scanf("%d%d",&a[i],&c[i]);
			s[i]=a[i];s[i+n]=c[i];
		}
		sort(s+1,s+1+2*n);
		m=1;pos[1]=s[1];
		for(i=2;i<=2*n;i++){
			if(s[i]!=s[i-1]){
				if(s[i]-s[i-1]>1){
					m++;pos[m]=s[i-1]+1;
				}
			    m++;pos[m]=s[i];
			}
		}
		build(1,m,1);
		for(i=1;i<=n;i++){
			l=find(a[i]);
			r=find(c[i]);
			update(l,r,i,1);
		}
		k=0;
		question(1);
		sort(sum+1,sum+1+k);
		num=1;
		for(i=2;i<=k;i++){
			if(sum[i]!=sum[i-1])num++;
		}
		printf("%d
",num);
	}
	return 0;
}