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POJ 2528 Mayor's posters

/*
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POJ 2528 Mayor's posters
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----問題描述：
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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.
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The city council has finally decided to build an electoral wall for placing the posters and introduce the following rules:
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Every candidate can place exactly one poster on the wall.
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All posters are of the same height equal to the height of the wall;
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the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown).
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The wall is divided into segments and the width of each segment is one byte.
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Each poster must completely cover a contiguous number of wall segments.
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They have built a wall 10000000 bytes long (such that there is enough place for all candidates).
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When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width.
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Moreover, the candidates started placing their posters on wall segments already occupied by other posters.
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Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections.
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Your task is to find the number of visible posters when all the posters are placed given the information about posters' size,
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their place and order of placement on the electoral wall.
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----輸入：
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The first line of input contains a number c giving the number of cases that follow.
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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.
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----輸出：
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For each input data set print the number of visible posters after all the posters are placed.
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----樣例輸入：
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1
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5
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1 4
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2 6
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8 10
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3 4
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7 10
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----樣例輸出：
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4
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----分析：
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線段樹。
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*/
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#include <iostream>
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#include <algorithm>
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using namespace std;
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template<unsigned int N>
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class CSegTree
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{
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public :
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void init( int b, int e ){
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init( 1, b, e );
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}
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void modify( int b, int e, int d ){
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begin = b;
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end = e;
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data = d;
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modify( 1 );
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}
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int query( void ){
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memset( visible, 0, sizeof( visible ) );
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query( 1 );
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data = 0;
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for( int i = 1; i < N; ++i ){
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if( visible[ i ] ){
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++data;
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}
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}
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return data;
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}
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private :
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void init( int node, int b, int e ){
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left[ node ] = b;
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right[ node ] = e;
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id[ node ] = 0;
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if( b < e ){
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init( node << 1, b, ( b + e ) >> 1 );
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init( ( node << 1 ) + 1, ( ( b + e ) >> 1 ) + 1, e );
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}
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}
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void modify( int node ){
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if( ( end < left[ node ] ) || ( right[ node ] < begin ) ){
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return;
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}
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if( data == id[ node ] ){
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return;
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}
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if( ( begin <= left[ node ] ) && ( right[ node ] <= end ) ){
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id[ node ] = data;
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return;
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}
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if( id[ node ] ){
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id[ node << 1 ] = id[ ( node << 1 ) + 1 ] = id[ node ];
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id[ node ] = 0;
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}
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modify( node << 1 );
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modify( ( node << 1 ) + 1 );
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if( id[ node << 1 ] == id[ ( node << 1 ) + 1 ] ){
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id[ node ] = id[ node << 1 ];
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}
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}
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void query( int node ){
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if( id[ node ] ){
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visible[ id[ node ] ] = true;
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return;
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}
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if( left[ node ] >= right[ node ] ){
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return;
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}
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query( node << 1 );
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query( ( node << 1 ) + 1 );
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}
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enum{ L = N * 3 };
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typedef int IA[ L ];
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IA left, right, id;
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bool visible[ N ];
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int begin, end, data;
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};
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template<unsigned int N, unsigned int NT>
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class CLine
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{
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public :
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friend istream & operator>>( istream & is, CLine<N,NT> & li ){
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is >> li.n;
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for( int i = 1; i <= li.n; ++i ){
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is >> li.left[ i ] >> li.right[ i ];
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}
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return is;
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}
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void init_tree( CSegTree<NT> & tree ){
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int i, j;
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n2 = n << 1;
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for( j = i = 1; i <= n; ++i,++j ){
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line[ j ].p = left[ i ];
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line[ j ].id = i;
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line[ j ].bLeft = true;
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++j;
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line[ j ].p = right[ i ];
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line[ j ].id = i;
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line[ j ].bLeft = false;
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}
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sort( line + 1, line + n2 + 1 );
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tp = 0;
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line[ 0 ].p = -123456;
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for( i = 1; i <= n2; ++i ){
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if( line[ i ].bLeft ){
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left[ line[ i ].id ] = ( line[ i - 1 ].p == line[ i ].p ? tp : ++tp );
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}
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else{
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right[ line[ i ].id ] = ( line[ i - 1 ].p == line[ i ].p ? tp : ++tp );
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}
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}
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tree.init( 1, tp );
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for( i = 1; i <= n; ++i ){
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tree.modify( left[ i ], right[ i ], i );
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}
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}
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private :
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struct SLine
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{
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bool operator<( const SLine & b ){
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return p < b.p;
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}
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int p, id;
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bool bLeft;
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};
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SLine line[ N * 2 ];
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int left[ N ], right[ N ], n, n2, tp;
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};
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const int L = 30009, TL = L * 2;
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CSegTree<TL> tree;
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CLine<L,TL> line;
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int main(){
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int td;
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cin >> td;
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while( td-- ){
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cin >> line;
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line.init_tree( tree );
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cout << tree.query() << endl;
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}
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return 0;
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}
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posted on 2012-04-22 22:50 coreBugZJ 閱讀(557) 評論(0) 編輯收藏引用所屬分類: ACM 、Algorithm 、DataStructure 、課內作業

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POJ 2528 Mayor's posters