Mayor's posters

Mayor's posters

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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

/*
题意：给横坐标轴染色，每次给一个区间内的染色，每染一次会把上一次的覆盖，经过n次染色只会，问你从最上面看，横坐标上总共有几种颜色

初步思路：区间染色问题，就是一个区间set问题，然后最后的查询的时候，记录一下就行了

#超内存：需要离散化
*/
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int n,l[10005],r[10005];
int t;
bool color[10005];//用于记录颜色
int res=0;
int X[20010];
int k;
int kk;
/****************************线段树基础模板*********************************/

const int maxn=20010+5;
#define lson i*2, l, m
#define rson i*2+1, m+1, r

struct Segtree{

    int setv[maxn<<2];

    void PushDown(int i)
    {
        if(setv[i]>0){
            setv[i*2]=setv[i*2+1]=setv[i];
            setv[i]=-1;//向下更新完了就没有继续用的意义了
        }
    }

    void build(int i,int l,int r)
    {
        // cout<<l<<" "<<r<<" "<<i<<endl;
        setv[i]=0;
        if(l==r)
            return ;
        int m=(l+r)>>1;
        build(lson);
        build(rson);
    }
    void query(int ql,int qr,int i,int l,int r)
    {
        if(ql<=l&&r<=qr){
            if(setv[i]>0){
                color[setv[i]]=true;
                return ;
            }
        } 
        if(l==r) return ;
        PushDown(i);
        int m=(l+r)>>1;
        if(ql<=m) query(ql,qr,lson);
        if(m<qr) query(ql,qr,rson);
    }

    void update(int ql,int qr,int val,int i,int l,int r)
    {
        // cout<<l<<" "<<r<<endl;
        if(ql<=l&&r<=qr)
        {
            setv[i]=val;
            return ;
        }
        PushDown(i);
        int m=(l+r)>>1;
        if(ql<=m) update(ql,qr,val,lson);
        if(m<qr) update(ql,qr,val,rson);
    }
};
Segtree segtree;
/****************************线段树基础模板*********************************/
void init(){
    memset(color,false,sizeof color);
    res=0;
    k=1;
    kk=1;
}
int findx(int key){
    int l=1,r=k,mid;
    while(l<=r){
        mid=(l+r)>>1;
        if(X[mid]==key)
            return mid;
        else if(X[mid]>key)
            r=mid-1;
        else l=mid+1;
    }
    return 0;
}
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&t);
    while(t--){
        init();
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d%d",&l[i],&r[i]);
            X[kk++]=l[i];
            X[kk++]=r[i];
        }
        sort(X+1,X+kk+1);
        for(int i=2;i<kk;i++){//离散化
            if(X[i]!=X[i-1])
                X[++k]=X[i];
        }
        segtree.build(1,1,k);
        for(int i=1;i<=n;i++){
            int L=findx(l[i]);
            int R=findx(r[i]);
            segtree.update(L,R,i,1,1,k);
        }
        segtree.query(1,k,1,1,k);
        for(int i=1;i<=n;i++){
            if(color[i]) res++;
        }
        printf("%d
",res);
    }
    return 0;
}

5
1 4
2 6
8 10
3 4
7 10

Sample Output

4