poj2528 Mayor's posters(线段树之成段更新)
2014-02-25 23:01
232 查看
Mayor's posters
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 37346
Accepted: 10864
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.
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 37346
Accepted: 10864
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.
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 const int maxn=11111; bool hash[maxn]; int li[maxn],ri[maxn]; int X[maxn*3]; int col[maxn<<2]; int cnt; void PushDown(int rt) { if(col[rt]!=-1) { col[rt<<1]=col[rt<<1|1]=col[rt]; col[rt]=-1; } } void update(int L,int R,int c,int l,int r,int rt) { if(L<=l&&R>=r) { col[rt]=c; return ; } PushDown(rt); int m=(l+r)>>1; if(L<=m) update(L,R,c,lson); if(m<R) update(L,R,c,rson); } void query(int l,int r,int rt) { if(col[rt]!=-1) { if(!hash[col[rt]]) cnt++; hash[col[rt]]=true; return ; } if(l==r) return ; int m=(l+r)>>1; query(lson); query(rson); } int Bin(int key,int n,int X[]) { int l=0,r=n-1; while(l<=r) { int m=(l+r)>>1; if(X[m]==key) return m; if(X[m]<key) l=m+1; else r=m-1; } return -1; } int main() { int T,n; scanf("%d",&T); while(T--) { scanf("%d",&n); int nn=0; for(int i=0;i<n;i++) { scanf("%d%d",&li[i],&ri[i]); X[nn++]=li[i]; X[nn++]=ri[i]; } sort(X,X+nn); int m=1; for(int i=1;i<nn;i++) if(X[i]!=X[i-1]) X[m++]=X[i]; for(int i=m-1;i>0;i--) { if(X[i]!=X[i-1]+1) X[m++]=X[i-1]+1; } sort(X,X+m); memset(col,-1,sizeof(col)); for(int i=0;i<n;i++) { int l=Bin(li[i],m,X); int r=Bin(ri[i],m,X); update(l,r,i,0,m-1,1); } cnt=0; memset(hash,false,sizeof(hash)); query(0,m-1,1); printf("%d\n",cnt); } }
相关文章推荐
- 第二章
- bork file encrypter
- Hive 随谈(五)– Hive 优化
- C++学习之运算符重载的总结
- Wix 安装部署(二)自定义安装界面和行为
- Wix 安装部署(一)同MSBuild 自动生成打包文件
- OpenMP编程总结表
- poj 1159 Palindrome (反串LCS 与 DP两种方法)
- C++学习之运算符重载的总结
- 第一个JAVA程序
- iOS应用本地化(Localizable)
- 九度1080解题报告
- 阿里云服务搭建
- LeetCode - Clone Graph
- JavaScript 检查日期是否在指定的范围
- VS2010配置Git时出现"尚未提供文件名"错误
- 开博说明
- Sharepoint网站创建自定义导航全记录
- 31-Linux-服务-Apache-虚拟主机
- hdu 3661 Assignments(水题的解法)