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POJ 3277 City Horizon

/*
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POJ 3277 City Horizon
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----問題描述：
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Farmer John has taken his cows on a trip to the city!
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As the sun sets, the cows gaze at the city horizon and observe the beautiful silhouettes formed by the rectangular buildings.
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The entire horizon is represented by a number line with N (1 ≤ N ≤ 40,000) buildings.
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Building i's silhouette has a base that spans locations Ai through Bi
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along the horizon (1 ≤ Ai < Bi ≤ 1,000,000,000)
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and has height Hi (1 ≤ Hi ≤ 1,000,000,000).
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Determine the area, in square units, of the aggregate silhouette formed by all N buildings.
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----輸入：
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Line 1: A single integer: N
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Lines 2..N+1: Input line i+1 describes building i with three space-separated integers: Ai, Bi, and Hi
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----輸出：
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Line 1: The total area, in square units, of the silhouettes formed by all N buildings
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----樣例輸入：
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4
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2 5 1
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9 10 4
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6 8 2
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4 6 3
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----樣例輸出：
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16
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----分析：
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線段樹+離散化。
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*/
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/******************************************************
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AC 766MS
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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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typedef long long Lint;
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template<class T, 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 loc[], int b, int e ){
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this->loc = loc;
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init( 1, b, e );
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}
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void modify( int b, int e, int h ){
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begin = b;
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end = e;
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hei = h;
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modify( 1 );
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}
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T query( void ){
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return query( 1 );
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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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height[ node ] = 0;
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if( b + 1 < e ){
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init( node << 1, b, ( b + e ) >> 1 );
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init( ( node << 1 ) + 1, ( b + e ) >> 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( ( begin <= left[ node ] ) && ( right[ node ] <= end ) ){
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height[ node ] = max( height[node], hei );
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return;
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}
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if ( left[ node ] + 1 >= right[ node ] ) {
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return;
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}
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height[ node << 1 ] = max( height[ node << 1 ], height[ node ] );
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height[ ( node << 1 ) + 1 ] = max( height[ ( node << 1 ) + 1 ], height[ node ] );
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height[ node ] = 0;
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modify( node << 1 );
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modify( ( node << 1 ) + 1 );
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}
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T query( int node ){
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if( height[ node ] ){
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return ((T)(height[node])) * (((T)(loc[right[node]])) - loc[left[node]]);
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}
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if( left[ node ] + 1 >= right[ node ] ){
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return 0;
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}
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return query( node << 1 ) + query( ( node << 1 ) + 1 );
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}
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typedef int IA[ N * 4 ];
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IA left, right, height;
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int *loc;
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int begin, end, hei;
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};
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template<class T, 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<T, 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 ] >> li.height[ i ];
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}
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return is;
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}
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void init_tree( CSegTree<T, 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 = 1;
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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 = 0;
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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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loc[ tp ] = line[ i ].p;
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}
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for ( i = 1; i <= n; ++i ) {
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line[ i ].p = height[ i ];
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line[ i ].id = left[ i ];
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line[ i ].bLeft = right[ i ];
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}
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sort( line+1, line+n+1 );
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for ( i = 1; i <= n; ++i ) {
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height[ i ] = line[ i ].p;
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left[ i ] = line[ i ].id;
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right[ i ] = line[ i ].bLeft;
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}
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tree.init( loc, 1, tp );
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for( i = 1; i <= n; ++i ){
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tree.modify( left[ i ], right[ i ], height[ 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, bLeft;
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};
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SLine line[ N * 2 ];
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int left[ N ], right[ N ], height[ N ], loc[ N * 2 ], n, n2, tp;
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};
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const int L = 40009, TL = L * 2;
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CSegTree<Lint, TL> tree;
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CLine<Lint, L, TL> line;
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int main(){
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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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return 0;
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}
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posted on 2012-04-22 22:52 coreBugZJ 閱讀(630) 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 3277 City Horizon