POJ 2201（修正版）

在前一个代码中，由于在寻找一段区间的最大最小值，用朴素的搜索办法，故造成效率低下。

后来，发现有个很好的RMQ的ST算法可以解决这个问题，果断使用之，

结果AC，用时2454ms

虽然花费时间更多，但证明了我之前的想法是对的，也让我学会了一个新算法ST算法！

#include <cstdio> #include <algorithm> #include <cassert> #include <cmath> using namespace std; const int MAX = 50010; typedef struct _Node { int k; int a; int index; }Node; typedef struct _Tree { int parent; int left; int right; }Tree; typedef struct _SN { int left; int right; int parent; bool isleft; }SN; Node data[MAX]; Tree tree[MAX]; int rmq[MAX][32]; SN stack[MAX]; void initrmq( int n ) { int len = floor( log( double(n) )/ log( 2.0 ) ); int mid; for( int i = 1 ; i <= n ; ++i ) rmq[i][0] = i; for( int i = 1 ; i <= len ; ++i ) for( int j = 1 ; j <= n ; ++j ) { rmq[j][i] = rmq[j][i-1]; mid = j + ( 1 <<( i - 1 ) ) ; if( mid <= n && data[rmq[j][i-1]].a > data[rmq[mid][i-1]].a ) rmq[j][i] = rmq[mid][i-1]; } } int getrmq( int left , int right ) { int len = right - left + 1; int mid = floor( log( double(len) ) / log(2.0) ); int mm1 = rmq[ left ][ mid ]; int mm2 = rmq[ right - ( 1 << mid ) + 1 ][ mid ]; if( data[mm1].a > data[mm2].a ) return mm2; else return mm1; } int getmin( int left , int right ) { return getrmq( left , right ); //int min = left; //for( int i = left + 1 ; i <= right ; ++i ) // if( data[i].a < data[min].a ) // min = i ; //return min; } void setchild( int parent , bool isleft , int value ) { if( isleft ) tree[ parent ].left = value; else tree[ parent ].right = value; } void createtree( int left , int right , int parent , bool isleft ) { int stacki = 0 ; stack[ stacki].left = left; stack[ stacki ].right = right; stack[ stacki ].parent = parent; stack[ stacki ].isleft = isleft; while( stacki != -1 ) { left = stack[ stacki ].left ; right = stack[ stacki ].right ; parent = stack[ stacki ].parent; isleft = stack[ stacki ].isleft; stacki--; if( left > right ) setchild( parent , isleft , 0 ); else if( left == right ) { tree[data[left].index].parent = parent; setchild( parent , isleft , data[left].index ); setchild( data[left].index , true , 0 ); setchild( data[left].index , false , 0 ); } else { int mid = getmin( left , right ); tree[ data[mid].index ].parent = parent; setchild( parent , isleft , data[mid].index ); stacki++; stack[ stacki ].left = left ; stack[ stacki ].right = mid - 1; stack[ stacki ].parent = data[ mid ].index; stack[ stacki ].isleft = true; stacki++; stack[ stacki ].left = mid + 1 ; stack[ stacki ].right = right; stack[ stacki ].parent = data[ mid ].index; stack[ stacki ].isleft = false; } } } bool cmp( const Node& left , const Node & right ) { return left.k < right.k ; } int main() { int n; scanf("%d",&n); for( int i = 1 ; i <= n ; ++i ) { scanf("%d%d",&data[i].k , &data[i].a ); data[i].index = i; } sort( &data[1] , &data[n+1] , cmp ); initrmq( n ); createtree( 1 , n , 0 , true ); printf("YES/n"); for( int i = 1 ; i <= n ; ++i ) printf("%d %d %d/n",tree[i].parent , tree[i].left , tree[i].right ); }