poj 3264 Balanced Lineup

purplest — Fri, 05 Aug 2011 15:17:00 GMT

果的RMQ的ST

#include <stdio.h>
#include <math.h>

int n, q;
int num[50050], dpmax[50050][20], dpmin[50050][20];

int max ( int a, int b)
{
    return a>b?a:b;
}
int min ( int a, int b)
{
    return a>b?b:a;
}

void creatmax()
{
    int i, j;
    for ( i = 1; i <= n; i++ )
        dpmax[i][0] = num[i];
    int k= log(n-1.0)/log(2.0);
    for ( j = 1; j <= k; j++)
        for ( i = 1 ; i+(1<<j)-1<=n; i++ )
            dpmax[i][j] = max( dpmax[i][j-1] , dpmax[i+(1<<(j-1))][j-1] );
}

void creatmin()
{
    int i, j;
    for ( i = 1; i <= n; i++ )
        dpmin[i][0] = num[i];
    int k= log(n-1.0)/log(2.0);
    for ( j = 1; j <= k; j++)
        for ( i = 1 ; i+(1<<j)-1<=n; i++ )
            dpmin[i][j] = min( dpmin[i][j-1] , dpmin[i+(1<<(j-1))][j-1] );
}

int getmax( int l, int r )
{
    int k= log(r-l+1.0)/log(2.0);
    return max(dpmax[l][k], dpmax[r-(1<<k)+1][k]);
}

int getmin( int l, int r )
{
    int k= log(r-l+1.0)/log(2.0);
    return min(dpmin[l][k], dpmin[r-(1<<k)+1][k]);
}

int main()
{
    scanf("%d%d", &n, &q);
    int i, j;
    for ( i = 1; i <= n; i++ )
    {
        scanf("%d", num+i);
    }
    creatmax();
    creatmin();
    int a, b;
    while (q--)
    {
        scanf("%d%d", &a, &b);
        printf("%d\n", getmax(a, b)-getmin(a, b));
    }
    return 0;
}

purplest 2011-08-05 23:17 发表评论

C++博客-purplest-随笔分类-RMQ

poj 3264 Balanced Lineup