| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 61616 | Accepted: 17812 |
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.
Sample Input
151 42 68 103 47 10Sample Output
4大意：有一面長度1e7的墻，然后告訴你張貼的海報的順序以及覆蓋的位置。讓你計算最后能看到的最多的海報數。用線段樹處理，剛接觸這東西，還不是很懂，然后需要離散化，因為最多10000個區(qū)間，20000個點，你不可能直接開1e7這么大的空間。
不知道為什么要開8倍空間。
按理說兩倍空間應該夠了，只能說是數據了吧。
作為線段樹的題應該說并不是很難，重點是離散化。
注意！！！離散化的時候，如果區(qū)間不相鄰的那么在編號的時候不要編成相鄰的。這個poj數據水，根本沒考慮過這樣的數據！！。
張貼海報，就是區(qū)間更新，我們從最后一張張貼的海報開始。
附上幾組數據：
531 101 36 1062 92 1714 1521 2315 1826 2635 64 56 831 101 36 1051 42 68 103 47 10正確答案自然是3 5 2 3 4.自己畫圖看看就知道了。附上正確的代碼：
//#include <bits/stdc++.h>#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int MAXN=1e4+7;int n,m;int postl[MAXN],postr[MAXN];struct node{ int l,r; bool iscover;}tree[MAXN<<4];int ha[10000005];int p[MAXN<<1];void build_tree(int i,int l,int r){ tree[i].l=l; tree[i].r=r; tree[i].iscover=0; if(l==r)return; int mid=(l+r)>>1; build_tree(i<<1,l,mid); build_tree(i<<1|1,mid+1,r);}bool post(int i,int l,int r)//貼上一張海報{ if(tree[i].iscover)return 0;//如果大區(qū)間已經被覆蓋 if(tree[i].l==l&&tree[i].r==r)//如果沒被覆蓋過 { tree[i].iscover=1; return 1; } int mid=(tree[i].l+tree[i].r)>>1; bool ans; if(r<=mid)ans=post(i<<1,l,r); else if(l>mid)ans=post(i<<1|1,l,r); else { int p1=post(i<<1,l,mid); int p2=post(i<<1|1,mid+1,r); ans=p1||p2; } //向上更新 if(tree[i<<1].iscover&&tree[i<<1|1].iscover)tree[i].iscover=1; return ans;}int main(){ int t; int cnt; scanf("%d",&t); while(t--) { cnt=0; scanf("%d",&n); for(int i=0;i<n;++i) { scanf("%d%d",&postl[i],&postr[i]); p[cnt++]=postl[i]; p[cnt++]=postr[i]; } sort(p,p+cnt); cnt=unique(p,p+cnt)-p; int pos=0; ha[p[0]]=0; for(int i=1;i<cnt;++i) { if(p[i]-p[i-1]==1)ha[p[i]]=++pos; else { pos+=2; ha[p[i]]=pos; } } build_tree(1,0,pos); int ans=0; for(int i=n-1;i>=0;--i) { if(post(1,ha[postl[i]],ha[postr[i]]))ans++; } printf("%d/n",ans); } return 0;}
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