BZOJ1636&&1699:[USACO2007JAN]Balanced Lineup
浅谈\(RMQ\):https://www.cnblogs.com/AKMer/p/10128219.html
题目传送门:https://lydsy.com/JudgeOnline/problem.php?id=1636
题目传送门:https://lydsy.com/JudgeOnline/problem.php?id=1699
裸的\(RMQ\)
时间复杂度:\(O(nlogn+m)\)
空间复杂度:\(O(nlogn)\)
代码如下:
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=5e4+5;
int n,m;
int a[maxn],Log[maxn];
int f[17][maxn],g[17][maxn];
int read() {
int x=0,f=1;char ch=getchar();
for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
return x*f;
}
void make_st() {
Log[0]=-1;
for(int i=1;i<=n;i++)
Log[i]=Log[i>>1]+1;
for(int i=1;i<17;i++)
for(int j=1;j+(1<<i)-1<=n;j++) {
f[i][j]=max(f[i-1][j],f[i-1][j+(1<<(i-1))]);
g[i][j]=min(g[i-1][j],g[i-1][j+(1<<(i-1))]);
}
}
int query(int l,int r) {
int x=Log[r-l+1];
int mx=max(f[x][l],f[x][r-(1<<x)+1]);
int mn=min(g[x][l],g[x][r-(1<<x)+1]);
return mx-mn;
}
int main() {
n=read(),m=read();
for(int i=1;i<=n;i++)
f[0][i]=g[0][i]=a[i]=read();
make_st();
for(int i=1;i<=m;i++) {
int l=read(),r=read();
printf("%d\n",query(l,r));
}
return 0;
}