poj 3264 Balanced Lineup

RMQ（區間最值問題）

題意：先給出一組數，然後給出區間，求區間內最大值與最小值的差

思路：RMQ的ST

#include<iostream>
#include<cmath>
using namespace std;
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define N 50010
int height[N];
int fmax[N][16],fmin[N][16],n; //二維數組的縱座標是 1<<j 是指數的冪 不用N那麼大
void rmqInit()
{
	int k = (int)(log((double)n)/log(2.0));
	int i,j;
	for(i = 0;i < n;i++)
	{
		fmax[i][0] = height[i];
		fmin[i][0] = height[i];
	}
	for(j = 1;j <= k;j++)  //先遍歷j 再遍歷i
		for(i = 0;i+(1<<j)-1 < n;i++)
		{
			int m = i + (1<<(j-1));
			fmax[i][j] = max(fmax[i][j-1],fmax[m][j-1]);
			fmin[i][j] = min(fmin[i][j-1],fmin[m][j-1]);
		}
}
int rmq(int a,int b)
{
	int k = (int)(log(double(b-a+1))/log(2.0));
	int ta = max(fmax[a][k],fmax[b-(1<<k)+1][k]);
	int tb = min(fmin[a][k],fmin[b-(1<<k)+1][k]);
	return ta - tb;
}
int main()
{
	int q;
	scanf("%d%d",&n,&q);
	int i;
	for(i = 0;i < n;i++)
		scanf("%d",height+i);
	rmqInit();
	int a,b;
	while(q--)
	{
		scanf("%d%d",&a,&b);
		printf("%d\n",rmq(a-1,b-1));
	}
	return 0;
}