RMQ算法

 题目链接：
 http://acm.nyist.net/JudgeOnline/problem.php?pid=119
 1 #include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;
#define maxn 100001

int ma[maxn][17], mi[maxn][17], N, Q;
//状态转移方程为：
//dp[j][i] = max(dp[j][i-1], dp[j+(1<<(i-1))][i-1]);
void rmq_init(){
    int i, j, t;
    int m = int(log(double(N))/log(2.0));
    for(i = 1; i <= m; ++i)
        for(j = 1; j <= N; ++j){
            t = j + (1 << (i - 1));
            if(t <= N){
                ma[j][i] = max(ma[j][i-1], ma[t][i-1]);
                mi[j][i] = min(mi[j][i-1], mi[t][i-1]);
            }
            else{
                ma[j][i] = ma[j][i-1];
                mi[j][i] = mi[j][i-1];
            }
        }
}

int query(int m, int n)
{
    int t = int(log(double(n - m + 1))/log(2.0));
    int a = max(ma[m][t], ma[n - (1<<t) + 1][t]);  //区间最大值
    int b = min(mi[m][t], mi[n - (1<<t) + 1][t]); //区间最小值
    return a - b;
}

int main()
{
    int i, m, n;
    while(scanf("%d%d",&N, &Q) != EOF){
        for(i = 1; i <= N; ++i){      //初值(dp[i][0])
            scanf("%d",&ma[i][0]);
            mi[i][0] = ma[i][0];
        }
        rmq_init();
        for(i = 0; i < Q; ++i)
        {
            scanf("%d%d",&m, &n);
            printf("%d
",query(m, n));
        }
    }
    return 0;
}