[洛谷P3865]【模板】ST表

题目大意：区间静态最大值

题解：ST表，zkw线段树

ST表：

st[i][j]存[i,i+$j^{2}$-1]的最大值，查询时把区间分成两个长度相同的小区间（可重复）

#include<cstdio>
#include<cctype>
using namespace std;
const int maxn=100010;
int n,m,M;
int st[maxn][20],lg[maxn];
inline int max(int a,int b){return a>b?a:b;}
inline void read(int &x){
	char t=getchar();
	while (!isdigit(t))t=getchar();
	for (x=t^48,t=getchar();isdigit(t);t=getchar())x=x*10+(t^48);
}
int main(){
	read(n),read(m);
	lg[0]=-1;
    for (int i=1;i<=n;i++)lg[i]=lg[i>>1]+1; 
    for (int i=1;i<=n;i++)read(st[i][0]);
    for (int i=1;i<=17;i++){
    	int tmp=1<<i-1;
    	for (int j=1;j+(tmp<<1)-1<=n;j++)st[j][i]=max(st[j][i-1],st[j+tmp][i-1]);
	}
    while (m--) {
    	int x,y,tmp;
        read(x),read(y);
        tmp=lg[y-x+1];
        printf("%d
",max(st[x][tmp],st[y-(1<<tmp)+1][tmp]));
    }
    return 0;
}

ZKW线段树：

#include<cstdio>
#include<cctype>
using namespace std;
const int maxn=100010;
int n,m,M;
int ts[maxn<<2];
inline int max(int a,int b){return a>b?a:b;}
inline void read(int &x){
	char t=getchar();
	while (!isdigit(t))t=getchar();
	for (x=t^48,t=getchar();isdigit(t);t=getchar())x=x*10+(t^48);
}
int ask(int s,int t){
	int ans=-2147483647;
	for (s+=M-1,t+=M+1;s^t^1;s>>=1,t>>=1){
		if (~s&1)ans=max(ans,ts[s^1]);
		if (t&1)ans=max(ans,ts[t^1]);
	}
	return ans;
}
int main(){
	read(n),read(m);
	for (M=1;M<=n+1;M<<=1);
	for (int i=M+1;i<=M+n;i++)read(ts[i]);
	for (int i=M-1;i;i--)ts[i]=max(ts[i<<1],ts[i<<1|1]);
	while (m--){
		int x,y;
		read(x),read(y);
		printf("%d
",ask(x,y));
	}
	return 0;
}