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[poj2528]Mayor‘s posters

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 66154		Accepted: 19104

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 題目大意：T組測試數據，每組測試數據有N張海報，按次序貼在板子上，我們可以將其抽象為一條直線，每張海報占據的區域[L,R]，問最後可以貼幾張海報。 試題分析：標記每個區間是否只有一種顏色，如果是的話訪問這個區間時看它的顏色編號有沒有被算進答案。更新時註意下傳標記。 　　　　　 POJ上的數據比較水，建議去試一試discuss中的數據，蠻良心的找出普通離散化的錯誤…… 　　　　　 比如說

1
3
1 3
6 10
1 10

//正確輸出：3
//錯誤輸出：2
//問題原因：離散化成了[1,2] [3,4] [1,4]，這樣確實只剩下2了

　　　　　　如何解決？在兩兩之差>1時(區域不會被完全覆蓋)，就可以在這裏插入一個節點以標記這裏有一個區間要算。

代碼：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c==‘-‘) f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-‘0‘;
	return x*f;
}
const int MAXN=200001;
const int INF=999999;
int N,M;
int T;
int A[MAXN*2],a[MAXN*2],b[MAXN*2];
int tr[MAXN*8+100];
int cnt,tmp3;
bool flag[MAXN*8+100];
bool Hash[MAXN*8+100];
int tmp;

void tage_lazy(int rt,int l,int r){
	if(flag[rt]){
		flag[rt*2]=flag[rt*2+1]=true;
		tr[rt*2]=tr[rt*2+1]=tr[rt];
		flag[rt]=false;
	}
	return ;
}
void add(int l,int r,int rt,int L,int R){
	if(L<=l&&R>=r){
		tr[rt]=cnt;
		flag[rt]=true;
		return ;
	}
	tage_lazy(rt,l,r);
	int mid=(l+r)>>1;
	if(mid<R) add(mid+1,r,rt*2+1,L,R);
	if(mid>=L) add(l,mid,rt*2,L,R);
	return ;
}
int Que(int l,int r,int rt,int L,int R){
	if(flag[rt]){
		if(!Hash[tr[rt]]){
			Hash[tr[rt]]=true;
			return 1;
		}
		else return 0;
	}
	if(l==r) return 0;
	int mid=(l+r)>>1;
	return Que(l,mid,rt*2,L,R)+Que(mid+1,r,rt*2+1,L,R);
}

int main(){
    T=read();
    while(T--){
	    memset(tr,0,sizeof(tr));
	    memset(flag,false,sizeof(flag));
	    memset(Hash,false,sizeof(Hash));
    	N=read();tmp=tmp3=0;
    	for(int i=1;i<=N;i++){
    		++tmp;A[tmp]=a[tmp]=read();
    		++tmp;A[tmp]=a[tmp]=read();
		}
		sort(a+1,a+tmp+1);
		int tmp2=0;int treef=tmp;
		for(int i=1;i<=tmp;i++) 
		    if(a[i]==a[i-1]) treef--;
		    else b[++tmp3]=a[i];
		int k=tmp3;
		for(int i=1;i<=k;i++)
		    if(b[i]>b[i-1]+1) b[++tmp3]=b[i-1]+1;
		sort(b+1,b+tmp3+1);
		treef=tmp3;
		for(int i=1;i<=N;i++){
			++cnt;
			int l=lower_bound(b+1,b+tmp3+1,A[tmp2+1])-b;
			int r=lower_bound(b+1,b+tmp3+1,A[tmp2+2])-b;
			add(1,treef,1,l,r);
			tmp2+=2;
		}
		printf("%d\n",Que(1,treef,1,1,treef)); 
	}
}

【線段樹】Mayor's posters