NYOJ 119 士兵杀敌 RMQ ST

Posted 2020-10-15 lemonsbiscuit

NYOJ 119 士兵杀敌（三） RMQ ST

题目链接：http://acm.nyist.net/JudgeOnline/problem.php?pid=119

思路：

ST在线 预处理O(nlogn) 查询O(1) 运行时间：828ms

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;
const int maxn = 100005;
const int maxp = 17;
int maxnum[maxn][maxp],minnum[maxn][maxp];
inline void init(int n) {
    for(int j=1; j<=17; ++j)
        for(int i=1; i<=n; ++i)
            if((i+(1<<j)-1)<=n) {
                maxnum[i][j]=max(maxnum[i][j-1],maxnum[i+(1<<(j-1))][j-1]);
                minnum[i][j]=min(minnum[i][j-1],minnum[i+(1<<(j-1))][j-1]);
            }
}
int main() {
    int n,m,l,r,k;
    scanf("%d %d",&n,&m);
    for(int i=1; i<=n; ++i) scanf("%d",&minnum[i][0]),maxnum[i][0]=minnum[i][0];
    init(n);
    for(int i=1; i<=m; ++i) {
        scanf("%d %d",&l,&r);
        k=(int)(log(r-l+1.0)/log(2.0));
        printf("%d\n", max(maxnum[l][k],maxnum[r-(1<<k)+1][k]) - min(minnum[l][k],minnum[r-(1<<k)+1][k]));
    }
    return 0;
}