poj 2019 二维RMQ-白红宇

poj 2019 二维RMQ

阅读量：6332 次

发布时间：2019-06-22

本文共 3003 字，大约阅读时间需要 10 分钟。

Cornfields

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 6169		Accepted: 3039

Description

FJ has decided to grow his own corn hybrid in order to help the cows make the best possible milk. To that end, he's looking to build the cornfield on the flattest piece of land he can find.

FJ has, at great expense, surveyed his square farm of N x N hectares (1 <= N <= 250). Each hectare has an integer elevation (0 <= elevation <= 250) associated with it.

FJ will present your program with the elevations and a set of K (1 <= K <= 100,000) queries of the form "in this B x B submatrix, what is the maximum and minimum elevation?". The integer B (1 <= B <= N) is the size of one edge of the square cornfield and is a constant for every inquiry. Help FJ find the best place to put his cornfield.

Input

* Line 1: Three space-separated integers: N, B, and K.

* Lines 2..N+1: Each line contains N space-separated integers. Line 2 represents row 1; line 3 represents row 2, etc. The first integer on each line represents column 1; the second integer represents column 2; etc.

* Lines N+2..N+K+1: Each line contains two space-separated integers representing a query. The first integer is the top row of the query; the second integer is the left column of the query. The integers are in the range 1..N-B+1.

Output

* Lines 1..K: A single integer per line representing the difference between the max and the min in each query.

Sample Input

5 3 15 1 2 6 31 3 5 2 77 2 4 6 19 9 8 6 50 6 9 3 91 2

Sample Output

/*poj 2019 二维RMQ前段时间不知道poj怎么了，现在补上二维RMQ简单，求解区间最大最小值的差hhh-2016-02-02 19:25:50*/#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            using namespace std;typedef long long ll;typedef long double ld;using namespace std;const int maxn = 255;int dp[maxn][maxn][8][8];int dp1[maxn][maxn][8][8];int tmap[maxn][maxn];int mm[maxn];void iniRMQ(int n,int m){ for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) { dp[i][j][0][0] = tmap[i][j]; dp1[i][j][0][0] = tmap[i][j]; } for(int ti = 0; ti <= mm[n]; ti++) for(int tj = 0; tj <= mm[m]; tj++) if(ti+tj) for(int i = 1; i+(1<
             
              <= n; i++) for(int j = 1; j+(1<
              
               <= m; j++) { if(ti) { dp[i][j][ti][tj] = max(dp[i][j][ti-1][tj],dp[i+(1<<(ti-1))][j][ti-1][tj]); dp1[i][j][ti][tj] = min(dp1[i][j][ti-1][tj],dp1[i+(1<<(ti-1))][j][ti-1][tj]); } else { dp[i][j][ti][tj] = max(dp[i][j][ti][tj-1],dp[i][j+(1<<(tj-1))][ti][tj-1]); dp1[i][j][ti][tj] = min(dp1[i][j][ti][tj-1],dp1[i][j+(1<<(tj-1))][ti][tj-1]); } }}int RMQ(int x1,int y1,int x2,int y2){ int k1 = mm[x2-x1+1]; int k2 = mm[y2-y1+1]; x2 = x2 - (1<
               
                < 255;i++) { mm[i] = (i&(i-1)) == 0 ? mm[i-1]+1:mm[i-1]; } while(scanf("%d%d%d",&n,&m,&k) != EOF) { for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) { scanf("%d",&tmap[i][j]); } iniRMQ(n,n); while(k--) { scanf("%d%d",&a,&b); printf("%d\n",RMQ(a,b,a+m-1,b+m-1)); } }}

转载于:https://www.cnblogs.com/Przz/p/5409630.html

你可能感兴趣的文章