hduoj 1754 I Hate it

I Hate it

对于区间的值进行修改与查询最大，采用线段树解决。用一个变量来记录某个区间的最大值，同时每次更新后，需要递归回调每个区间的子区间的最大值。用于更新父节点的最大值。程序如下:

/*ID: csuchenanPROG: hduoj 1754 I hate itLANG: C++*/#include<stdio.h>#include<string.h>#include<stdlib.h>#define max(a , b) (a) > (b) ? (a) : (b)#define MAXN 200005#define MY_MAX -1struct Node{int left  ;int right ;int nmax  ;}node[MAXN * 4] ;void build(int left , int right , int layer) ;void query(int left , int right , int layer , int &nmax) ;void update(int pos , int nval  , int layer) ;void insert(int pos , int nval  , int layer) ;int n ;int m ;void work() ;int main(int argc , char * argv[]){while(scanf("%d %d" , &n , &m)!=EOF)work() ;return 0 ;}void work(){//build the treebuild(1 , n , 1) ;int i ;int data ;i = 1 ;while(i <= n){scanf("%d" , &data)  ;//insert the data into the treeinsert(i , data , 1) ;i++ ;}char c ;i = 0 ;int p ;int q ;int nmax  ;while(i < m){getchar() ;scanf("%c %d %d" , &c , &p , &q) ;if(c == 'Q'){nmax = MY_MAX ;query(p , q , 1 , nmax) ;printf("%d\n" , nmax)   ;}else{update(p , q , 1) ;}i ++ ;}}void build(int left , int right , int layer ){node[layer].left = left ;node[layer].right = right ;node[layer].nmax = MY_MAX ;if(left == right){return  ;}int mid = (left + right) / 2 ;int llayer = layer * 2 ;int rlayer = layer * 2 + 1 ;build(left , mid , llayer) ;build(mid + 1 , right , rlayer) ;return ;}void insert(int pos , int nval , int layer){if(node[layer].nmax < nval){node[layer].nmax = nval ;}if(pos == node[layer].left && pos == node[layer].right){return ;}int mid = (node[layer].left + node[layer].right) / 2 ;int nlayer ;if(mid >= pos){nlayer = layer * 2 ;insert(pos , nval , nlayer) ;return ;}nlayer = layer * 2 + 1 ;insert(pos , nval , nlayer) ;return ;}void update(int pos , int nval , int layer){if(node[layer].left == pos && node[layer].right == pos){node[layer].nmax = nval ;return ;}int mid = (node[layer].left + node[layer].right) / 2 ;int llayer ;int rlayer ;llayer = layer * 2 ;rlayer = layer * 2 + 1 ;if(pos <= mid){update(pos , nval , llayer) ;}else{update(pos , nval , rlayer) ;}node[layer].nmax = max(node[llayer].nmax , node[rlayer].nmax) ;return ;}void query(int left , int right , int layer , int &nmax){if(node[layer].left == left && node[layer].right == right){nmax = max( node[layer].nmax , nmax );return ;}int mid = (node[layer].left + node[layer].right) / 2 ;int nlayer ;if(right <= mid){nlayer = layer * 2 ;query(left , right , nlayer , nmax ) ;return ;}if(left > mid){nlayer = layer * 2 + 1 ;query(left , right , nlayer , nmax) ;return ;}nlayer = layer * 2 ;query(left , mid , nlayer , nmax) ;query(mid + 1 , right , nlayer + 1 , nmax) ;return ;}