cogs 2554. [福利]可持久化线段树

题目链接

cogs 2554. [福利]可持久化线段树

题解

没有

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
 
inline int read () {
	int x = 0,f = 1;
	char c = getchar();
	while(c < '0' || c > '9'){if(c=='-')f=-1; c=getchar();}
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar();
	return x*f;
}
const int maxn = 100007;
int n,m,a[maxn],tot = 1,root[maxn];
struct Chairman_Tree {
	int ch[2],num ;
} ;
Chairman_Tree t[maxn << 4];
#define lc t[x].ch[0]
#define rc t[x].ch[1]
inline void update(int x) { t[x].num = std::max(t[lc].num,t[rc].num); }
void build(int & x,int l,int r) {
	x = ++ tot;
	if(l == r) {
		t[x].num = a[l];return ;
	}
	int mid = l + r >> 1;
	build(lc,l,mid); build(rc,mid+1,r);return ;
	update(x) ;
}
 
int query(int x,int l,int r,int L,int R) {
	if(l == r) return t[x].num; 
	int mid = l + r >>1; 
	int ret = 0; 
	if(mid >= L) ret = std::max(ret,query(lc,l,mid,L,R)) ; 
	if(mid < R) ret = std::max(ret,query(rc,mid+1,r,L,R)) ; 
	return ret ;
}
 
void modify(int pre,int &x,int l,int r,int pos,int w) {
	t[x = ++tot] = t[pre]; 
	if(l == r) { 
		t[x].num = w;return ; 
	} 
	int mid = l + r >> 1; 
	if(pos <= mid) modify(t[pre].ch[0],lc,l,mid,pos,w); 
	else modify(t[pre].ch[1],rc,mid + 1,r,pos,w); 
	update(x); 
} 
int main () {
	freopen("longterm_segtree.in","r",stdin);
	freopen("longterm_segtree.out","w",stdout);
	n = read(),m = read(); int tmp = 1; 
	for(int i = 1;i <= n;++ i) a[i] = read(); 
	build(root[1],1,n); 
	for(int T,op,a,b,i = 1;i <= m;++ i) { 
		op = read(),T = read(); 
		if(op == 1) { 
			a = read(),b = read(); 
			modify(root[T],root[++tmp],1,n,a,b); 
		} 
		else { 
			a = read(),b = read();//,root[i] = root[T]; 
			printf("%d
",query(root[T],1,n,a,b)); 
		} 
	} 
	return 0; 
}
原文地址:https://www.cnblogs.com/sssy/p/8710463.html