Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 64939		Accepted: 18770

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. 

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.

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. 

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

4

题意：
N个人（1<=N<=10000）依次贴N张等高的海报，
给出每张海报的左端点，右端点li，ri（1<=li<=ri<=10000000）。
后面的海报可能会把前面的海报盖住。
问最后能看见几张海报。

题解：
一个比较明显的区间染色问题。
暴力肯定会TLE+爆内存，没有问题吧。
对我这种蒟蒻来说好像只会用刚学的线段树。
但是端点范围太大，直接开数组明显不可行。
所以还要离散化一下。
所谓的离散化，在我理解看来
就是将一个很大的区间映射成一个很小的区间，
而不改变原有的覆盖关系。
但是对于这道题而言，简单的离散化可能会出现错误。
比如说：
例子一:[1,10][1,4][5,10]
例子二:[1,10][1,4][6,10]
它们普通离散化后都变成了[1,4][1,2][3,4]。
线段2覆盖了[1,2],线段3覆盖了[3,4],那么线段1是否被覆盖掉了呢?
例子一是被覆盖掉了,而例子二没有被覆盖。
解决的办法，就是在距离大于1的两个相邻节点间插入一个点，保证准确性。

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
struct node{int l,r;}a[100001]; //节点信息 
int N,ans;
int c[400001]; //区间每个点的颜色 
bool vis[100001]; //记录颜色访问信息 
int x[100001]; //离散化后的区间 
inline int read(){
    int x=0,f=1;
    char c=getchar();
    for(;!isdigit(c);c=getchar())
        if(c==‘-‘)
            f=-1;
    for(;isdigit(c);c=getchar())
        x=x*10+c-‘0‘;
    return x*f;
}
void cover(int num){
    if(c[num]!=-1){
        c[num<<1]=c[num<<1|1]=c[num];
        c[num]=-1;
    }
}
void update(int L,int R,int col,int l,int r,int num){
    //如果这个点在需要染色的区间内，直接更新并且返回 
    if(l>=L && r<=R){
        c[num]=col;
        return;
    }
    cover(num); //更新
    int mid=(l+r)>>1; 
    if(L<=mid) update(L,R,col,l,mid,num<<1); //更新左儿子 
    if(R>mid) update(L,R,col,mid+1,r,num<<1|1); //更新右儿子 
}
void query(int l,int r,int num){
    //累加未出现的颜色 
    if(c[num]!=-1){
        if(!vis[c[num]]) ans++;
        vis[c[num]]=1;
        return;
    }
    if(l==r) return; //如果到了叶子则返回 
    int mid=(l+r)>>1;
    query(l,mid,num<<1); //统计左儿子 
    query(mid+1,r,num<<1|1); //统计右儿子
} 
int main(){
    int T=read();
    while(T--){
        memset(c,-1,sizeof(c));
        memset(vis,0,sizeof(vis));
        N=read();ans=0;
        int p=0; //输入点数
        int q=1; //离散化后点数 
        for(int i=0;i<N;i++){
            a[i].l=read();a[i].r=read();
            x[p++]=a[i].l;x[p++]=a[i].r; 
        }
        sort(x,x+p);
        //去重操作 
        for(int i=1;i<p;i++) 
            if(x[i]!=x[i-1])
                x[q++]=x[i];
        //离散化操作
        for(int i=q-1;i>=1;i--)
            if(x[i]>x[i-1]+1)
                x[q++]=x[i-1]+1;
        sort(x,x+q);
        //染色操作
        for(int i=0;i<N;i++){
            int l=lower_bound(x,x+q,a[i].l)-x;
            int r=lower_bound(x,x+q,a[i].r)-x;
            update(l,r,i,0,q,1);
        }
        //统计操作 
        query(0,q,1);
        printf("%d\n",ans);
    }
    return 0;
}