RMQ模板题 poj3264 Balanced Lineup

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poj3264 ST算法模板题

#include <iostream>#include <cstdio>using namespace std;const int maxn = 80000;const int maxm = 30;int d_min[maxn][maxm],d_max[maxn][maxm],a[maxn];int n;void RMQ_init(){    int i,j;    for(i = 1; i <= n; i++){        d_min[i][0] = a[i];        d_max[i][0] = a[i];    }    for(j = 1; (1<<j) <= n; j++)    for(i = 1; i + j - 1 <= n; i++){        d_min[i][j] = min(d_min[i][j-1],d_min[i + (1<<(j-1))][j-1]);        d_max[i][j] = max(d_max[i][j-1],d_max[i + (1<<(j-1))][j-1]);    }}int RMQ_min(int l,int r){    int k = 0;    while((1<<(k+1)) <= r-l+1)        k++;    return min(d_min[l][k], d_min[r-(1<<k)+1][k]);}int RMQ_max(int l,int r){    int k = 0;    while((1<<(k+1)) <= r-l+1)        k++;    return max(d_max[l][k], d_max[r-(1<<k)+1][k]);}int main(){    int q,l,r,i;    scanf("%d%d",&n,&q);    for(i = 1; i <= n; i++)    scanf("%d",&a[i]);    RMQ_init();    while(q--){        scanf("%d%d",&l,&r);        printf("%d\n",RMQ_max(l,r)-RMQ_min(l,r));    }    return 0;}

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