Cornfields(poj2019)

时间：2016-10-08 16:27:05 阅读：196 评论：0 收藏：0 [点我收藏+]

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Cornfields

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 6798		Accepted: 3315

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

Sample Output

5
思路：单调栈；
因为正方形的大小是固定的，然后我们先每列每个元素用单调队列维护最大最小值，然后在用维护的矩阵，在每行每个元素维护最大最小值
这样ama[i][j]就是以（i-b+1，j-b+1）为左上角，所有元素的最大值;同理ami[i][j]为最小。复杂度O（n*n）；

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<queue>
  6 #include<deque>
  7 #include<stack>
  8 #include<math.h>
  9 using namespace std;
 10 typedef long long LL;
 11 int ma[300][300];
 12 int maxx[300][300];
 13 int minn[300][300];
 14 int que[300*2];
 15 int ama[300][300];
 16 int ami[300][300];
 17 void get_maxx(int n,int k);
 18 void get_minn(int n,int k);
 19 int main(void)
 20 {
 21         int n,b,k;
 22         int i,j;
 23         while(scanf("%d %d %d",&n,&b,&k)!=EOF)
 24         {
 25                 for(i = 1; i <= n; i++)
 26                 {
 27                         for(j = 1; j <= n; j++)
 28                         {
 29                                 scanf("%d",&ma[i][j]);
 30                         }
 31                 }
 32                 get_minn(n,b);
 33                 get_maxx(n,b);
 34                 while(k--)
 35                 {
 36                         int x;
 37                         int y;
 38                         scanf("%d %d",&x,&y);
 39                         x+=b-1;
 40                         y+=b-1;
 41                         printf("%d\n",ama[x][y]-ami[x][y]);
 42                 }
 43         }
 44         return 0;
 45 }
 46 void get_maxx(int n,int k)
 47 {
 48         int i,j;
 49         for(j = 1; j <= n; j++)
 50         {
 51                 int head = 1;
 52                 int rail = 0;
 53                 for(i = 1; i <= n; i++)
 54                 {
 55                         if(head > rail)
 56                         {
 57                                 que[++rail] = i;
 58                         }
 59                         else
 60                         {
 61                                 int id = que[rail];
 62                                 while(ma[id][j] <= ma[i][j])
 63                                 {
 64                                         rail--;
 65                                         if(rail < head)
 66                                                 break;
 67                                         id = que[rail];
 68                                 }
 69                                 que[++rail] = i;
 70                         }
 71                         int ic = que[head];
 72                         while(ic < max(0,i-k)+1)
 73                         {
 74                                 head++;
 75                                 ic = que[head];
 76                         }
 77                         maxx[i][j] = ma[que[head]][j];
 78                 }
 79         }
 80         for(i = 1; i <= n; i++)
 81         {
 82                 int head = 1;
 83                 int rail = 0;
 84                 for(j = 1; j <= n; j++)
 85                 {
 86                         if(head > rail)
 87                         {
 88                                 que[++rail] = j;
 89                         }
 90                         else
 91                         {
 92                                 int id = que[rail];
 93                                 while(maxx[i][id] <= maxx[i][j])
 94                                 {
 95                                         rail--;
 96                                         if(rail < head)
 97                                                 break;
 98                                         id = que[rail];
 99                                 }
100                                 que[++rail] = j;
101                         }
102                         int ic = que[head];
103                         while(ic < max(0,j-k)+1)
104                         {
105                                 head++;
106                                 ic = que[head];
107                         }
108                         ama[i][j] = maxx[i][que[head]];
109                 }
110         }
111 }
112 void get_minn(int n,int k)
113 {
114         int i,j;
115         for(j = 1; j <= n; j++)
116         {
117                 int head = 1;
118                 int rail = 0;
119                 for(i = 1; i <= n; i++)
120                 {
121                         if(head > rail)
122                         {
123                                 que[++rail] = i;
124                         }
125                         else
126                         {
127                                 int id = que[rail];
128                                 while(ma[id][j] >= ma[i][j])
129                                 {
130                                         rail--;
131                                         if(rail < head)
132                                                 break;
133                                         id = que[rail];
134                                 }
135                                 que[++rail] = i;
136                         }
137                         int ic = que[head];
138                         while(ic < max(0,i-k)+1)
139                         {
140                                 head++;
141                                 ic = que[head];
142                         }
143                         minn[i][j] = ma[que[head]][j];
144                 }
145         }
146         for(i = 1; i <= n; i++)
147         {
148                 int head = 1;
149                 int rail = 0;
150                 for(j = 1; j <= n; j++)
151                 {
152                         if(head > rail)
153                         {
154                                 que[++rail] = j;
155                         }
156                         else
157                         {
158                                 int id = que[rail];
159                                 while(minn[i][id] >= minn[i][j])
160                                 {
161                                         rail--;
162                                         if(rail < head)
163                                                 break;
164                                         id = que[rail];
165                                 }
166                                 que[++rail] = j;
167                         }
168                         int ic = que[head];
169                         while(ic < max(0,j-k)+1)
170                         {
171                                 head++;
172                                 ic = que[head];
173                         }
174                         ami[i][j] = minn[i][que[head]];
175                 }
176         }
177 }

Cornfields(poj2019)

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原文地址：http://www.cnblogs.com/zzuli2sjy/p/5939011.html

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