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最新推荐文章于 2022-03-05 17:21:40 发布

转载最新推荐文章于 2022-03-05 17:21:40 发布 · 57 阅读

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原文链接：http://www.cnblogs.com/jie-dcai/p/4296437.html

本文探讨了RMQ算法的难点及其优化方法，通过对比两种不同的代码实现方式，详细解释了如何通过改进数据结构和算法逻辑来提高效率，并提供了解决问题的实例代码。

简单的RMQ，可我怎么写都WA。不明白，找了一个和我相似的贴过了，要赶着去外婆家。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stack>
#define eps 1e-5
#define MAXN 255
#define MAXM 111111
#define INF 1000000000
using namespace std;
int mx[MAXN][MAXN][8], mi[MAXN][MAXN][8];
int n, b, q, a, c;
void rmqinit()
{
    int l = int(log(double(n)) / log(2.0)) ;
    for(int k = 1; k <= n ; k++)
        for(int j = 1; j <= l; j++)
            for(int i = 1; i + (1 << (j - 1))- 1 <= n; i++)
            {
                mx[k][i][j] = max(mx[k][i][j - 1], mx[k][i + (1 << (j - 1 ))][j - 1]) ;
                mi[k][i][j] = min(mi[k][i][j - 1], mi[k][i + (1 << (j - 1 ))][j - 1]) ;
            }
}
int rmqmax(int lx, int ly, int rx, int ry) // lx, ly为左上角的点 rx ry为右下角的点
{
    int l = int(log(double(ry - ly + 1)) / log(2.0));
    int ret = -1;
    for(int k = lx; k <= rx ; k++)
        ret = max(ret, max(mx[k][ly][l], mx[k][ry - (1 << l) + 1][l]));
    return ret;
}
int rmqmin(int lx, int ly, int rx, int ry) // lx, ly为左上角的点 rx ry为右下角的点
{
    int l = int(log(double(ry - ly + 1)) / log(2.0));
    int ret = INF;
    for(int k = lx; k <= rx ; k++)
        ret = min(ret, min(mi[k][ly][l], mi[k][ry - (1 << l) + 1][l]));
    return ret;
}
int main()
{
    while(scanf("%d%d%d", &n, &b, &q) != EOF)
    {
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
            {
                scanf("%d",&a);
                mx[i][j][0] = mi[i][j][0] = a ;
            }
        rmqinit();
        while(q--)
        {
            scanf("%d%d", &a, &c) ;
            int rx = a + b - 1;
            if(rx > n) rx = n;
            int ry = c + b - 1;
            if(ry > n) ry = n;
            printf("%d\n", rmqmax(a, c, rx, ry) - rmqmin(a, c, rx, ry)) ;
        }
    }
    return 0;
}

MINE：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string.h>
#include <queue>
#include <cmath>
#include <vector>
using namespace std;

int num[255][255];
int f1[255][255][30];
int f2[255][255][30];
int n,b,q;
const int inf=1000000000;

int rmq_max(int p,int i, int j) { 
     int k = (int)(log(double(j-i+1)) / log(2.0)), t1;
     t1 = max(f1[p][i][k], f1[p][j - (1<<k) + 1][k]);
     return t1;
}
int rmq_min(int p,int i, int j) { 
     int k = (int)(log(double(j-i+1)) / log(2.0)), t2;
     t2 = min(f2[p][i][k], f2[p][j - (1<<k) + 1][k]);
     return t2;
}

int main(){
	int l,u;
	while(scanf("%d%d%d",&n,&b,&q)!=EOF){
		int k = (int) (log((double)n) / log(2.0)); 
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++)
			scanf("%d",&num[i][j]);
 		    for(int j = 0; j < n; j++) {
            	f1[i][j][0] = num[i][j]; 
            	f2[i][j][0] = num[i][j];
			}
			for(int p = 1; p <= k; p++) {
				for(int t = 0; t + (1 << p) - 1 < n; t++)   {
				 	int	m = t + (1 << (p - 1)); 
					f1[i][t][p] = max(f1[i][t][p-1], f1[i][m][p-1]);
					f2[i][t][p] = min(f2[i][t][p-1], f2[i][m][p-1]);
				}
			}
		}
		int maxn=-1,minn=inf;
		for(int i=1;i<=q;i++){
			scanf("%d%d",&u,&l);
			l--;u--;
			for(int p=u;p<(u+b);p++){
				maxn=max(maxn,rmq_max(p,l,l+b-1));
				minn=min(minn,rmq_min(p,l,l+b-1));
			}
			printf("%d\n",maxn-minn);
		}
	}
	return 0;
}

转载于:https://www.cnblogs.com/jie-dcai/p/4296437.html