RMQ DP

最原始的DP，空間和時間複雜度O(n^2)。在線

#include "stdio.h"
#include "string.h"
#define M 100

#define MIN(a, b) ((a)<(b)?(a):(b))

int num[M] = {2, 1, 67, 33, 53, 21, 6, 35, 44, 61};
int n = 10;

int dp[M][M];

void DP(){
	int i, j;
	memset(dp, 0, sizeof(dp));
	for(i=0; i<n; i++)
		dp[i][i] = num[i];
	for(j=1; j<n; j++){
		for(i=0; i+j<n; i++)
			dp[i][i+j] = MIN(dp[i][i+j-1], dp[i+1][i+j]);
	}
}

void main(){
	int i, j;
	DP();
	for(j=1; j<n; j++){
		for(i=0; i+j<n; i++)
			printf("%d-%d:%d ", i, i+j, dp[i][i+j]);
		printf("\n");
	}
}

轉換一下，j表示從i開始，長度爲j的這段區間，時間空間複雜度和上面相同。在線

#include "stdio.h"
#include "string.h"
#define M 100

#define MIN(a, b) ((a)<(b)?(a):(b))

int num[M] = {2, 1, 67, 33, 53, 21, 6, 35, 44, 61};
int n = 10;

int dp[M][M];

void DP(){
	int i, j;
	memset(dp, 0, sizeof(dp));
	for(i=0; i<n; i++)
		dp[0][i] = num[i];
	for(j=1; j<n; j++){
		for(i=0; i+j<n; i++)
			dp[j][i] = MIN(dp[j-1][i], dp[j-1][i+1]);
	}
}

void main(){
	int i, j;
	DP();
	for(j=1; j<n; j++){
		for(i=0; i+j<n; i++)
			printf("%d-%d:%d ", i, i+j, dp[j][i]);
		printf("\n");
	}
}

滾動數組，空間複雜度降到O(n)，時間複雜度沒減少。離線

#include "stdio.h"
#include "string.h"
#define M 100

#define MIN(a, b) ((a)<(b)?(a):(b))

int num[M] = {2, 1, 67, 33, 53, 21, 6, 35, 44, 61};
int n = 10;

int dp[2][M];

void DP(){
	int i, j;
	int p, q;
	memset(dp, 0, sizeof(dp));
	p = 0; q = 1;
	for(i=0; i<n; i++)
		dp[p][i] = num[i];
	for(j=1; j<n; j++){
		for(i=0; i+j<n; i++){
			dp[q][i] = MIN(dp[p][i], dp[p][i+1]);
			printf("%d-%d:%d ", i, i+j, dp[q][i]);
		}
		p = (p+1)%2; q = (q+1)%2;
		printf("\n");
	}
}

void main(){
	DP();
}

還是用稀疏表（ST）好一些吧