Mayor's posters-POJ2528(线段树+离散化)
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.
样例输入
1 5 1 4 2 6 8 10 3 4 7 10
样例输出
4
离散化 的大概思路 : 比如说给你一组 数据 1 4 1000 100000， 如果直接
开线段, 显然是浪费, 那么我们只要 进行 映射 :
1 1
4 2
1000 3
100000 4
接下来 我们只要对 1 2 3 4 建立线段树就行了 只需要
[1,4]的区间
离散化就相当于是先做映射，然后再建树。
本题大意：给定一些海报，可能相互重叠，告诉你每个海报的宽度（高度都一样的）和先后叠放顺序，问没有被完全盖住的有多少张？
海报最多10000张，但是墙有10000000块瓷砖长，海报不会落在瓷砖中间。
如果直接建树，就算不TLE,也会MLE。即单位区间长度太多。
其实10000张海报，有20000个点，最多有19999个区间。对各个区间编号，就是离散化。然后建数。
其实浮点数也是一样离散化的。
还有好多需要注意的地方。这题的线段树要四倍的，普通的三倍不行了。
细节决定成败：
#include<stdio.h> #include<math.h> #include<algorithm> #include<iostream> #include<string.h> #include<stdlib.h> using namespace std; #define N 40005 #define Lson r<<1 #define Rson r<<1|1 int aa[N]; struct node { int L,R; bool is; int mid() { return (L+R)/2; } }a[N*4]; struct Node{int x,y;}p[N]; void BuildTree(int r,int L,int R) { a[r].L=L; a[r].R=R; a[r].is=false; if(L==R) return; BuildTree(Lson,L,a[r].mid()); BuildTree(Rson,a[r].mid()+1,R); } void Up(int r) { if(a[r].L!=a[r].R) { if(a[Lson].is && a[Rson].is) a[r].is=true; } } bool Update(int r,int L,int R) { if(a[r].is==true) return false; if(a[r].L==L && a[r].R==R) { a[r].is=true; return true; } bool sum; if(R<=a[r].mid()) sum=Update(Lson,L,R); else if(L>a[r].mid()) sum=Update(Rson,L,R); else { bool a1=Update(Lson,L,a[r].mid()); bool a2=Update(Rson,a[r].mid()+1,R); sum=a2+a1; } Up(r); return sum; } int main() { int T,n,k=0,i; scanf("%d",&T); while(T--) { int ans=0; k=0; scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%d %d",&p[i].x,&p[i].y); aa[k++]=p[i].x; aa[k++]=p[i].x-1; aa[k++]=p[i].y; aa[k++]=p[i].y+1; } sort(aa,aa+k); int len=unique(aa,aa+k)-aa; BuildTree(1,1,len); for(i=n;i>0;i--) { int l=lower_bound(aa,aa+len,p[i].x)-aa; int r=lower_bound(aa,aa+len,p[i].y)-aa; if(Update(1,l,r)==true) ans++; } printf("%d\n",ans); } return 0; }
