描述

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. 

输入

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.

输出

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. 
http://acm.pku.edu.cn/JudgeOnline/images/2528_1.jpg

样例输入

151 42 68 103 47 10

样例输出

4

来源

POJ

上传者

iphxer

　　离散化+线段树。

　　难度较高的一道线段树的题，用了离散化处理。在南阳理工学院的oj（http://acm.nyist.net/JudgeOnline/problem.php?pid=9）上提交成功，但是在poj上提交WA，不知道为什么……听说poj上测试数据有些问题，还要算上端点两侧的点？？不清楚，有知道的同仁告诉我一下，拜谢。。。

　　题意：

　　按顺序给你一些海报，这些海报可能相互重叠，求最后没有被完全盖住的海报的数量。

　　给你海报的两个端点的位置，最多有10000张海报，但是位置最多可以是10000000。

　　思路：

　　由于涉及到区间，首先应该想到线段树，但是海报的宽度太大，如果给你一个1-10000000大小的海报怎么办？肯定MLE超内存啊（这的new多少节点）。所以就要用到离散化的方法，听起来很神秘，但理解起来很简单，例如给你两张海报，[1,6]和[3,10]，排序之后是1,3,6,10，离散化为1,2,3,4，即第一个区间为[1,3]，第二个区间为[2,4]，如此，创建一个[1,4]的线段树即可，而不用创建[1,10]的线段树，节省了空间。

　　代码：

 1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 #include <algorithm> 5 using namespace std; 6  7 #define MAXN 20000 8 bool tree[MAXN*4+1];    //存储区间有无海报 9 short hash[10000010];    //离散化后的端点位置10 short x[MAXN+10];    //存储海报的两个端点11 struct post{12     short L;13     short R;14 }a[MAXN+10];    //原数组15 int cnt;    //记录露在外面的海报数量16 17 bool Insert(int d,int L,int R,int l,int r)18 {19     if(tree[d])20         return false;21     if(L==l && R==r){    //找到区间22         tree[d] = true;23         return true;24     }25 26     int mid  = (L+R)/2;27     bool res;28     if(mid>=r){29         res =  Insert(d<<1,L,mid,l,r);30     }31     else if(mid<l){32         res =  Insert(d<<1|1,mid+1,R,l,r);33     }34     else {35         bool f1 = Insert(d<<1,L,mid,l,mid);36         bool f2 = Insert(d<<1|1,mid+1,R,mid+1,r);37         res =  f1||f2;38     }39     if(tree[d<<1] && tree[d<<1|1])40         tree[d] = true;41     return res;42 }43 44 int main()45 {46     int T;47     scanf("%d",&T);48     while(T--){49         memset(tree,0,sizeof(tree));50         memset(hash,0,sizeof(hash));51         memset(x,0,sizeof(x));52         int i,n,nCount=0;53 54         //输入55         scanf("%d",&n);56         for(i=0;i<n;i++){57             scanf("%d%d",&a[i].L,&a[i].R);58             x[nCount++] = a[i].L;59             x[nCount++] = a[i].R;60         }61 62         sort(x,x+nCount);63         nCount = unique(x,x+nCount) - x;    //去重64 65         int key=1;66         for(i=0;i<nCount;i++){    //离散化处理67             hash[x[i]] = key;68             if(i<nCount-1){69                 if(x[i+1]-x[i]==1)70                     key++;71                 else    72                     key=key+2;73             }74         }75 76         cnt = 0;77         for(i=n-1;i>=0;i--){78             if(Insert(1,1,key,hash[a[i].L],hash[a[i].R]))    //将离散化后的海报插入到线段树中79                 cnt++;80         }81         printf("%d\n",cnt);82     }83     return 0;84 }

Freecode : www.cnblogs.com/yym2013