Mayor's posters

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 56864		Accepted: 16445

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

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

using namespace std;

#define N 1005

int m,ans,li[N],ri[N];

int x[N<<],col[N<<],hash[N];

void pushdown(int rt){

    col[rt<<]=col[rt<<|]=col[rt];

    col[rt]=-;

}

void updata(int rt,int l,int r,int val,int x,int y){

    if(x>=l&&y<=r){

        col[rt]=val;return ;

    }

    if(col[rt]!=-) pushdown(rt);

    int mid=(x+y)>>;

    if(mid>=l) updata(rt<<,l,r,val,x,mid);

    if(mid<r) updata(rt<<|,l,r,val,mid+,y);

}

void query(int rt,int l,int r){

    if(l==r){

        if(!hash[col[rt]]){

            hash[col[rt]]=;

            ans++;

        }

        return ;

    }

    if(col[rt]!=-) pushdown(rt);

    int mid=(l+r)>>;

    query(rt<<,l,mid);

    query(rt<<|,mid+,r);

}

int search(int ll,int hh,int xx){

    while(ll<=hh){

        int mid=(ll+hh)>>;

        if(x[mid]==xx) return mid;

        else if(x[mid]>xx) hh=mid-;

        else ll=mid+;

    }

    return -;

}

int main(){

    int t,n;

    scanf("%d",&t);

    while(t--){

        memset(col,-,sizeof col);

        memset(hash,,sizeof hash);

        int tot();

        scanf("%d",&n);

        for(int i=;i<=n;i++){

            scanf("%d%d",&li[i],&ri[i]);

            x[++tot]=li[i];

            x[++tot]=ri[i];

        }

        sort(x+,x+tot+);

        m=;

        for(int i=;i<=tot;i++){

            if(x[i]!=x[i-]) x[++m]=x[i];

        }

        for(int i=m;i>;i--){

            if(x[i]-x[i-]>) x[++m]=x[i]-;

        }

        sort(x+,x+m+);

        for(int i=;i<=n;i++){

            int l=search(,m,li[i]);

            int r=search(,m,ri[i]);

            updata(,l,r,i,,m);

        }

        ans=;

        query(,,m);

        printf("%d\n",ans);

    }

    return ;

}

#include<iostream>

#include<cstring>

#include<cstdio>

#include<algorithm>

using namespace std;

#define N 10005

int m,li[N],ri[N];

int x[N<<],col[N<<],ans;

bool hash[N];

void pushdown(int rt) {

     col[rt<<]=col[rt<<|]=col[rt];

     col[rt]=-;

}

void updata(int rt,int L,int R,int c,int l,int r){

     if(l>=L&&r<=R){

         col[rt]=c;return ;

     }

     if(col[rt]!=-) pushdown(rt);

     int mid=(l+r)>>;

     if(mid>=L) updata(rt<<,L,R,c,l,mid);

     if(mid<R)  updata(rt<<|,L,R,c,mid+,r);

}

void query(int rt,int l,int r){

    if(l==r){

        if(!hash[col[rt]]){

            hash[col[rt]]=;

            ans++;

        }

       return;

    }

    if(col[rt]!=-) pushdown(rt);

    int mid=(l+r)>>;

    query(rt<<,l,mid);

    query(rt<<|,mid+,r);

}

int search(int ll,int hh,int xx){

    int mid;

    while (ll<=hh){

        mid=(ll+hh)>>;

        if(x[mid]==xx) return mid;

        else if(x[mid]>xx)  hh=mid-;

        else ll=mid+;

    }

    return -;

}

int main(){

    int t,n,i;

    scanf("%d",&t);

    while(t--){

        memset(col,-,sizeof(col));

        memset(hash,,sizeof(hash));

        int tot=;

        scanf("%d",&n);

        for(i=;i<=n;i++){

             scanf ("%d %d",&li[i],&ri[i]);

             x[++tot]=li[i];

             x[++tot]=ri[i];

        }

        sort(x+,x+tot+);

        m=;

        for(i=;i<=tot;i++){

             if(x[i]!=x[i-]) x[++m]=x[i];

        }

        for(i=m;i>;i--){

            if(x[i]-x[i-]>) x[++m]=x[i]-;

        }

        sort(x+,x+m+);

        for(i=; i<=n; i++){

            int l=search(,m,li[i]);

            int r=search(,m,ri[i]);

            updata(,l,r,i,,m);

        }

        ans=;

        query(,,m);

        printf("%d\n",ans);

    }

    return ;

}