poj 3264 Balanced Lineup（RMQ）

来源：互联网发布：新人网络男主播技巧编辑：程序博客网时间：2024/06/05 20:00

题目地址

题目大意：给出N个数（1 ≤ N ≤ 50,000），Q个查询（1 ≤ Q ≤ 200,000），求区间的最大值与最小值的差值

解题思路：直接RMQ

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <queue>#include <string>#include <map>#include <stack>#include <list>#include <set>using namespace std;const int maxn = 50005;int FMAX[maxn][20], FMIN[maxn][20];void RMQ(int n){    for(int j = 1; j != 20; ++j)    {        for(int i = 1; i <= n; ++i)        {            if(i + (1 << j) - 1 <= n)            {                FMAX[i][j] = max(FMAX[i][j - 1], FMAX[i + (1 << (j - 1))][j - 1]);                FMIN[i][j] = min(FMIN[i][j - 1], FMIN[i + (1 << (j - 1))][j - 1]);            }        }    }}int main(){    int num, query;    int a, b;    while(scanf("%d %d", &num, &query) != EOF)    {        for(int i = 1; i <= num; ++i)        {            scanf("%d", &FMAX[i][0]);            FMIN[i][0] = FMAX[i][0];        }        RMQ(num);        while(query--)        {            scanf("%d%d", &a, &b);            int k = (int)(log(b - a + 1.0) / log(2.0));            int maxsum = max(FMAX[a][k], FMAX[b - (1 << k) + 1][k]);            int minsum = min(FMIN[a][k], FMIN[b - (1 << k) + 1][k]);            printf("%d\n", maxsum - minsum);        }    }    return 0;}

0 0