Poj 2528-Mayor's posters 线段切割

 

题目：http://poj.org/problem?id=2528

Mayor's posters

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 55156		Accepted: 16000

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

Source

Alberta Collegiate Programming Contest 2003.10.18

 

 

题意：给出一些海报，后贴的会把先贴的覆盖，问最多能看到多少张海报。

题解：

线段切割

学习了一个线段切割

其实很简单的。

具体看 IOI2004年薛矛的论文《剖析线段树与矩形切割》。

概括起来就是前面的线段不会影响后面的线段，所以从后面往前面推。对于每个线段，我们只考虑后面的线段对它的影响，影响有五类（论文第26页中有），我们可以将其简化为三类 :

不相交：

1.两个线段不相交。

相交：

2.前一个线段的左端点小于后一个线段的左端点（包括了 前一个线段覆盖了后一个线段的前半部分 和 前一个线段覆盖了后一个线段）

3.前一个线段的右端点大于后一个线段的右端点（包括了 前一个线段覆盖了后一个线段的后半部分 和 前一个线段覆盖了后一个线段）

具体看程序:

 #include<bits/stdc++.h>
 using namespace std;
 int N,x[],y[],ans[];
 int read()
 {
     int s=,fh=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')fh=-;ch=getchar();}
     while(ch>=''&&ch<=''){s=s*+(ch-'');ch=getchar();}
     return s*fh;
 }
 void Cover(int l,int r,int k,int k1)
 {
     while(k<=N&&(r<x[k]||l>y[k]))k++;
     if(k>=N+){ans[k1]+=(r-l+);return;}
     if(l<x[k])Cover(l,x[k]-,k+,k1);
     if(r>y[k])Cover(y[k]+,r,k+,k1);
 }
 int main()
 {
     int T,i,tot;
     T=read();
     while(T--)
     {
         N=read();
         for(i=;i<=N;i++){x[i]=read();y[i]=read();}
         memset(ans,,sizeof(ans));
         for(i=N;i>=;i--)Cover(x[i],y[i],i+,i);
         tot=;
         for(i=;i<=N;i++)if(ans[i]>)tot++;
         printf("%d\n",tot);
     }
     fclose(stdin);
     fclose(stdout);
     return ;
 }