POJ 2528 Mayor's posters(线段树,区间覆盖,单点查询)-白红宇

POJ 2528 Mayor's posters(线段树,区间覆盖,单点查询)

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发布时间：2019-06-19

本文共 4902 字，大约阅读时间需要 16 分钟。

Mayor's posters

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 45703		Accepted: 13239

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

151 42 68 103 47 10

Sample Output

用 col[rt] 来记录如今是什么颜色， -1 表示这个区间有多个颜色..

区间的范围比较大，先对它进行离散一下。

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                 using namespace std;#define root 1,n,1#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define lr rt<<1#define rr rt<<1|1typedef long long LL;const int oo = 1e9+7;const double PI = acos(-1.0);const double eps = 1e-6 ;const int N = 20010;const int mod = 2333333;int n , m ;LL col[N<<2] , tot ;bool tag[N<<2] ;void Down( int l , int r , int rt ) { if( col[rt] != -1 ) col[lr] = col[rr] = col[rt] ;}void Up( int rt ){ if( col[lr] == col[rr] && col[lr] != -1 ) col[rt] = col[lr]; else col[rt] = -1 ;}void build( int l , int r , int rt ){ col[rt] = -1 ; if( l == r ) return ; int mid = (l+r)>>1; build(lson),build(rson);}void update( int l , int r , int rt , int L , int R , int c ) {// cout << L <<' ' << R << ' ' << l << ' ' << r << endl ; if( L == l && r == R ) { col[rt] = c ; return ; } if( col[rt] == c ) return ; else Down( l,r,rt ) , col[rt] = -1 ; int mid = (l+r)>>1; if( R <= mid ) update(lson,L,R,c); else if( L > mid ) update(rson,L,R,c); else update(lson,L,mid,c) , update(rson,mid+1,R,c); Up(rt);}LL query( int l , int r , int rt , int x ) { if( col[rt] != -1 ) return col[rt]; Down(l,r,rt); int mid = (l+r)>>1; if( x <= mid ) return query(lson,x); else return query(rson,x);}struct node { int num , new_num , id ;}e[N];bool cmp1( const node &a , const node &b ) { return a.num < b.num ; }bool cmp2( const node &a , const node &b ) { return a.id < b.id ; }int main(){ #ifdef LOCAL freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout); #endif // LOCAL int _ ; scanf("%d",&_); while( _-- ) { memset( tag , false , sizeof tag ); tot = 0 ; scanf("%d",&n); for( int i = 0 ; i < n ; ++i ) { scanf("%d",&e[tot].num); e[tot].id = tot ; tot++; scanf("%d",&e[tot].num); e[tot].id = tot ; tot++; } sort( e , e + tot , cmp1 ); e[0].new_num = 1 ; for( int i = 1 ; i < tot ; ++i ) e[i].new_num = (e[i].num == e[i-1].num?e[i-1].new_num:e[i-1].new_num+1); sort( e , e + tot , cmp2 ); n = tot+10; build(root); for( int i = 0 ; i < tot ; i+=2 ){// cout << e[i].new_num << ' ' << e[i+1].new_num << endl ; update(root,e[i].new_num,e[i+1].new_num,i+1); }// cout << "ok" << endl ; for( int i = 1 ; i <= n ; ++i ) {// cout << query(root,i) << endl ; int color = query(root,i) ; if( color == -1 ) continue ; tag[color] = true ; } int ans = 0 ; for( int i = 1 ; i <= n ; ++i ) if(tag[i]) ans++; printf("%d\n",ans); }}

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转载于:https://www.cnblogs.com/hlmark/p/4248098.html

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