poj3277 City Horizon(离散化+线段树)

题意 : Farmer 带着他的牛在城市中散步。日落时，牛望着远方地平线上N( N <= 40000 )个建筑的背影。。问题是这N个建筑的背影所占的面积是多少?

其中一个建筑这样表示(A, B, H)，A B(1 =< A < B <= 10^10 )代表起点坐标与终点坐标, H( 1 =< H <= 10^10 )代表高度

 

覆盖问题，用线段树解决。但坐标系范围太大，建筑数相对较少，这会使坐标系内较空旷，因此本题要离散化

PS:通过本题学到了什么

     1.对于线段树，后来的更新可能会覆盖前面的更新

     2.计算面积时中间结果可能会溢出(注意)

#include <iostream>#include <algorithm>using namespace std;const int Max = 40000;// 对象struct Shadow {int b, e, h;}s[Max];struct TreeNode {int l, r;int c;}tree[2*Max*4];int p[2*Max]; // 离散化向量long long area;// 函数bool cmp_for_s( const Shadow & s1, const Shadow & s2 ){return s1.h < s2.h;}//-----------------离散化查找-------------------//int BS( int l, int r, const int key ){while( l <= r ) {int mid = ( l + r ) >> 1;if( key == p[mid] )return mid + 1;else if( key < p[mid] ) r = mid - 1;elsel = mid + 1;}return 0; // 未找到}//-----------------线段树操作-------------------//void Build( const int l, const int r, const int id ){tree[id].l = l;tree[id].r = r;tree[id].c = 0;if( r - l == 1 ) return;int mid = ( l + r ) >> 1;Build( l, mid, id * 2 );Build( mid, r, id * 2 + 1 );}void Update( const int l, const int r, const int h, const int id ){if( tree[id].l == l && tree[id].r == r ) {tree[id].c = h;return;}if( tree[id].c > 0 ) {tree[ id*2 ].c = tree[ id*2+1 ].c = tree[id].c;tree[id].c = 0;}int mid = ( tree[id].l + tree[id].r ) >> 1;if( r <= mid ) Update( l, r, h, id * 2 ); // 注意啊, 注意啊 // 开始写成了 Update( l, mid, ... )else if( l >= mid ) Update( l, r, h, id * 2 + 1 ); // 同样注意else {Update( l, mid, h, id * 2 );Update( mid, r, h, id * 2 + 1 );}}void Query( const int id ){if( tree[id].c > 0 ) { // 该死的中间结果溢出，测试时已经发现不对劲了。。area += ( p[ tree[id].r-1 ] - p[ tree[id].l-1 ] ) * ( long long )tree[id].c;return;}if( tree[id].r - tree[id].l == 1 ) return;Query( id * 2 );Query( id * 2 + 1 );}//-------------------------end---------------------------//int main(){int N;scanf( "%d", & N );int len = 0;for( int i = 0; i < N; ++ i ) {scanf( "%d%d%d", & s[i].b, & s[i].e, & s[i].h );p[ len++ ] = s[i].b;p[ len++ ] = s[i].e;}// 离散化::sort( p, p + 2 * N );::sort( s, s + N, ::cmp_for_s );// 建树::Build( 1, 2 * N, 1 );// 覆盖for( int i = 0; i < N; ++ i ) {int l = ::BS( 0, 2 * N - 1, s[i].b );int r = ::BS( 0, 2 * N - 1, s[i].e );Update( l, r, s[i].h, 1 );/*Show( 1 );*/}// 遍历树area = 0;::Query( 1 );printf( "%lld/n", area );system( "pause" );return 0;}