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.
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.
* 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 1 5 1 2 6 3 1 3 5 2 7 7 2 4 6 1 9 9 8 6 5 0 6 9 3 9 1 2
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 ",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 }