POJ 2201

静态建树有两种方法，一是每次找出根节点是什么，递归建树即可，二是假设原来的是空树，一个一个地插入去。

本题是后者，

先贴一下前者代码，超时了，但感觉此做法是对的

#include <cstdio>#include <algorithm>#include <cassert>using namespace std;const int MAX = 50010;typedef struct _Node{intk;inta;intindex;}Node;typedef struct _Tree{intparent;intleft;intright;}Tree;Nodedata[MAX];Treetree[MAX];intgetmin( int left , int right ){int min = left;for( int i = left + 1 ; i <= right ; ++i )if( data[i].a < data[min].a )min = i ;return min;}voidsetchild( int parent , bool isleft , int value ){if( isleft )tree[ parent ].left = value;elsetree[ parent ].right = value;}voidcreatetree( int left , int right , int parent , bool isleft ){if( left > right )setchild( parent , isleft , 0 );else if( left == right ){tree[data[left].index].parent = parent;setchild( parent , isleft , data[left].index );createtree( left , left - 1 , data[left].index , true );createtree( left , left - 1 , data[left].index , false );}else{intmid = getmin( left , right );tree[ data[mid].index ].parent = parent;setchild( parent , isleft , data[mid].index );createtree( left , mid - 1 , data[mid].index , true );createtree( mid + 1 , right , data[mid].index , false );}}boolcmp( 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 );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 );} 

下面是后者的代码，时间为901ms

#include <cstdio>#include <cstdlib>using namespace std;const int MAX = 50010;typedef struct _Node{int k , a , index;}Node;typedef struct _Tree{int parent , left , right ;}Tree;Nodedata[MAX];Nodestack[MAX];Treetree[MAX];int cmp( const void* left , const void* right ){Node*Left = ( Node* )left;Node*Right = ( Node * )right;return Left->k - Right->k;}int main(){int n , i , stacki;scanf("%d",&n);for( i = 1 ; i <= n ; ++i )scanf("%d%d",&data[i].k , &data[i].a ) , data[i].index = i ;qsort( &data[1] , n , sizeof( Node ) , cmp );stacki = -1;for( int j = 1 ; j <= n ; ++j ){for( i = stacki ; i >= 0 && stack[i].a > data[j].a ; --i );tree[ data[j].index ].parent = 0 ;tree[ data[j].index ].left = 0 ;tree[ data[j].index ].right = 0 ;if( i >= 0 ){tree[ stack[i].index ].right = data[j].index;tree[ data[j].index ].parent = stack[i].index;}if( i + 1 <= stacki ){tree[ stack[i+1].index ].parent = data[j].index ;tree[ data[j].index ].left = stack[i+1].index;}stacki = i + 1;stack[ stacki ] = data[j];}printf("YES/n");for( int i = 1 ; i <= n ; ++i ){printf("%d %d %d/n",tree[i].parent , tree[i].left , tree[i].right );}}