POJ 4047 Garden

POJ 4047 Garden

线段树题目，很简单。需要预处理一下，树的节点中存储的元素为区间的两个端点，然后该段中最大的值，还有延迟标记。预处理的时候先求出所有长度为k的序列的长度，然后建树。具体的程序如下：开始的时候忽略了测试例子数，所以一直runtime error。

/*ID : csuchenanPEOG: D GardenLANG: C++*/#include<stdio.h>#include<string.h>#include<algorithm>#define MAXN 201000using namespace std ;struct Node{int l ;int r ;int flag ;int nmax ;}node[4*MAXN] ; int nsum[MAXN] ;int nval[MAXN] ;int n ;int m ;int k ;bool init() ;void work() ;void pushdown(int root) ;void pushup(int root) ;void solve(int x , int y) ;void build(int l , int r , int c) ;void update(int l , int r , int val , int c) ;long long query(int l , int r , int c);int main(){int t ;scanf("%d" , &t) ;//freopen("out.txt" , "w" , stdout) ;while(t--){init() ;work() ;}return 0 ;}bool init(){if(scanf("%d %d %d" , &n , &m , &k)!=3){return 0 ;}memset(nsum , 0 , sizeof(nsum)) ;int i ;int j ;i = 1 ;while(i <= n){scanf("%d" , &nval[i]);i++ ;}for(i = 1 ; i <= k ; i ++)nsum[1] += nval[i] ;j = 2 ;for(i = k + 1 ; i <= n ; i ++){nsum[j] = nsum[j-1] + nval[i] - nval[i-k] ;j ++ ;}build(1 , n - k + 1 , 1) ;return 1 ;}void work(){int x ;int y ;int p ;int l ;int r ;while(m--){scanf("%d %d %d" , &p , &x , &y) ;if(p==0){solve(x , y - nval[x]) ;nval[x] = y ;}else if(p==1){solve(x , nval[y] - nval[x]) ;solve(y , nval[x] - nval[y]) ;int temp = nval[x] ;nval[x] = nval[y]  ;nval[y] = temp ;}else{l = x ;r = y - k + 1 ; printf("%d\n", query(l , r , 1)) ;} }}void solve(int x , int y){int l ;int r ;l = x - k + 1 ;if(l < 1)l = 1 ;r = x ;if(x + k - 1 > n)r = n - k + 1 ;update(l , r , y , 1) ;}void build(int l , int r , int c){node[c].l = l ;node[c].r = r ;node[c].flag = 0 ;if(l == r){node[c].nmax = nsum[l] ;return ;}int mid = (l + r)/2 ;build(l , mid , c * 2) ;build(mid + 1 , r , c * 2 + 1) ;pushup(c) ;}void pushup(int root){node[root].nmax = max(node[root * 2].nmax , node[root * 2 + 1].nmax) ;}void pushdown(int root){if(node[root].flag==0)return ;node[root*2].flag += node[root].flag ;node[root*2+1].flag += node[root].flag ;node[root*2].nmax += node[root].flag ;node[root*2+1].nmax += node[root].flag ;node[root].flag = 0 ;}void update(int l , int r , int val , int c){pushdown(c) ;if(l == node[c].l  && node[c].r == r){node[c].flag += val ;node[c].nmax += val ;return ;}//pushdown(c) ;int mid = (node[c].l + node[c].r)/2 ;if(r <= mid){update(l , r , val , c*2) ;}else if(mid < l){update(l , r , val , c*2+1) ;}else{update(l , mid , val , c*2) ;update(mid + 1 , r , val , c*2+1) ;}pushup(c) ; }long long query(int l , int r , int c){pushdown(c) ;if(l == node[c].l && r == node[c].r){return node[c].nmax ;}int mid = (node[c].l  + node[c].r)/2 ;if(r <= mid){return query(l , r , c * 2) ;}if(l > mid){return query(l , r , c * 2 + 1) ;}return max(query(l , mid , c * 2 ) , query(mid + 1 , r , c * 2 + 1)) ;}