Mayor's posters

Time Limit: 1000MS

 

Memory Limit: 65536K

Total Submissions: 32200

 

Accepted: 9347

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 li 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 <= li <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered li, li+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. 

POJ 2528 Mayor

Sample Input


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


Sample Output


4


Source

​​Alberta Collegiate Programming Contest 2003.10.18​​

题意：

在x轴贴海报，后贴的可能会把前面贴的给覆盖掉，问最后有多少不同的海报是能看到的。

思路：
坐标离散化+线段树

本题关键是海报要倒着贴，最后一个肯定是要显示出来的，然后我们用线段树去维护该区间是否被贴过，贴过，直接下一步，未被贴过，ans++

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

#define L(x) (x<<1)
#define R(x) (x<<1|1)

const int N=100010;     

int flag;

struct Tree{
    int l,r;
    bool vis;   //vis:这块区域是否完全被覆盖
}tree[N<<2];

struct node{
    int id,x;
}post[N<<2];

int cmp1(node a,node b){
    return a.x<b.x;
}

int cmp2(node a,node b){
    if(a.id==b.id)
        return a.x<b.x;
    return a.id>b.id;
}

void PushUp(int x){
    tree[x].vis=tree[L(x)].vis && tree[R(x)].vis;
}

void build(int L,int R,int x){
    tree[x].l=L;
    tree[x].r=R;
    tree[x].vis=0;
    if(tree[x].l==tree[x].r)
        return ;
    int mid=(L+R)>>1;
    build(L,mid,L(x));
    build(mid+1,R,R(x));
}

void query(int L,int R,int x){
    if(tree[x].vis==1)
        return ;
    if(tree[x].l>=L && tree[x].r<=R){
        tree[x].vis=1;
        flag=1;
        return ;
    }
    int mid=(tree[x].l+tree[x].r)>>1;
    if(L<=mid) query(L,R,L(x));
    if(R>mid)  query(L,R,R(x));
    PushUp(x);
}

int main(){
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<2*n;i+=2)
        {
            scanf("%d%d",&post[i].x,&post[i+1].x);
            post[i].id=post[i+1].id=i;
        }
        sort(post,post+2*n,cmp1);
        int tot=0,pre=0;
        for(int i=0;i<2*n;i++)
        {     
            if(post[i].x==pre)
                post[i].x=tot;
            else
            {
                pre=post[i].x;
                post[i].x=++tot;
            }
        }
        build(1,2*n,1);
        sort(post,post+2*n,cmp2);  ///倒着贴
        int ans=0;
        for(int i=0;i<2*n;i+=2)
        {
            int l=post[i].x;
            int r=post[i+1].x;
            flag=0;
            query(l,r,1);
            if(flag)
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}