Mosaic
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 213 Accepted Submission(s): 50
Problem Description
The God of sheep decides to pixelate some pictures (i.e., change them into pictures with mosaic). Here's how he is gonna make it: for each picture, he divides the picture into n x n cells, where each cell is assigned a color value. Then he chooses a cell, and checks the color values in the L x L region whose center is at this specific cell. Assuming the maximum and minimum color values in the region is A and B respectively, he will replace the color value in the chosen cell with floor((A + B) / 2).
Can you help the God of sheep?
Can you help the God of sheep?
Input
The first line contains an integer T (T ≤ 5) indicating the number of test cases. Then T test cases follow.
Each test case begins with an integer n (5 < n < 800). Then the following n rows describe the picture to pixelate, where each row has n integers representing the original color values. The j-th integer in the i-th row is the color value of cell (i, j) of the picture. Color values are nonnegative integers and will not exceed 1,000,000,000 (10^9).
After the description of the picture, there is an integer Q (Q ≤ 100000 (10^5)), indicating the number of mosaics.
Then Q actions follow: the i-th row gives the i-th replacement made by the God of sheep: xi, yi, Li (1 ≤ xi, yi ≤ n, 1 ≤ Li < 10000, Li is odd). This means the God of sheep will change the color value in (xi, yi) (located at row xi and column yi) according to the Li x Li region as described above. For example, an query (2, 3, 3) means changing the color value of the cell at the second row and the third column according to region (1, 2) (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4). Notice that if the region is not entirely inside the picture, only cells that are both in the region and the picture are considered.
Note that the God of sheep will do the replacement one by one in the order given in the input.
Each test case begins with an integer n (5 < n < 800). Then the following n rows describe the picture to pixelate, where each row has n integers representing the original color values. The j-th integer in the i-th row is the color value of cell (i, j) of the picture. Color values are nonnegative integers and will not exceed 1,000,000,000 (10^9).
After the description of the picture, there is an integer Q (Q ≤ 100000 (10^5)), indicating the number of mosaics.
Then Q actions follow: the i-th row gives the i-th replacement made by the God of sheep: xi, yi, Li (1 ≤ xi, yi ≤ n, 1 ≤ Li < 10000, Li is odd). This means the God of sheep will change the color value in (xi, yi) (located at row xi and column yi) according to the Li x Li region as described above. For example, an query (2, 3, 3) means changing the color value of the cell at the second row and the third column according to region (1, 2) (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4). Notice that if the region is not entirely inside the picture, only cells that are both in the region and the picture are considered.
Note that the God of sheep will do the replacement one by one in the order given in the input.
Output
For each test case, print a line "Case #t:"(without quotes, t means the index of the test case) at the beginning.
For each action, print the new color value of the updated cell.
For each action, print the new color value of the updated cell.
Sample Input
1
3
1 2 3
4 5 6
7 8 9
5
2 2 1
3 2 3
1 1 3
1 2 3
2 2 3
Sample Output
Case #1:
5
6
3
4
6
Source
二维线段树的水题了。
对于二维的矩阵，需要查询一个区域的最大和最小值。
修改单个点的值。
二维线段树直接搞，主要是修改的时候，更新操作要往两个方向进行。
和一维差不多，就是更新不同。
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2014/5/13 23:21:07 4 File Name :E:2014ACM专题学习数据结构二维线段树HDU4819.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 const int INF = 0x3f3f3f3f; 21 const int MAXN = 1010; 22 struct Nodey 23 { 24 int l,r; 25 int Max,Min; 26 }; 27 int locy[MAXN],locx[MAXN]; 28 struct Nodex 29 { 30 int l,r; 31 Nodey sty[MAXN*4]; 32 void build(int i,int _l,int _r) 33 { 34 sty[i].l = _l; 35 sty[i].r = _r; 36 sty[i].Max = -INF; 37 sty[i].Min = INF; 38 if(_l == _r) 39 { 40 locy[_l] = i; 41 return; 42 } 43 int mid = (_l + _r)/2; 44 build(i<<1,_l,mid); 45 build((i<<1)|1,mid+1,_r); 46 } 47 int queryMin(int i,int _l,int _r) 48 { 49 if(sty[i].l == _l && sty[i].r == _r) 50 return sty[i].Min; 51 int mid = (sty[i].l + sty[i].r)/2; 52 if(_r <= mid)return queryMin(i<<1,_l,_r); 53 else if(_l > mid)return queryMin((i<<1)|1,_l,_r); 54 else return min(queryMin(i<<1,_l,mid),queryMin((i<<1)|1,mid+1,_r)); 55 } 56 int queryMax(int i,int _l,int _r) 57 { 58 if(sty[i].l == _l && sty[i].r == _r) 59 return sty[i].Max; 60 int mid = (sty[i].l + sty[i].r)/2; 61 if(_r <= mid)return queryMax(i<<1,_l,_r); 62 else if(_l > mid)return queryMax((i<<1)|1,_l,_r); 63 else return max(queryMax(i<<1,_l,mid),queryMax((i<<1)|1,mid+1,_r)); 64 } 65 }stx[MAXN*4]; 66 int n; 67 void build(int i,int l,int r) 68 { 69 stx[i].l = l; 70 stx[i].r = r; 71 stx[i].build(1,1,n); 72 if(l == r) 73 { 74 locx[l] = i; 75 return; 76 } 77 int mid = (l+r)/2; 78 build(i<<1,l,mid); 79 build((i<<1)|1,mid+1,r); 80 } 81 //修改值 82 void Modify(int x,int y,int val) 83 { 84 int tx = locx[x]; 85 int ty = locy[y]; 86 stx[tx].sty[ty].Min = stx[tx].sty[ty].Max = val; 87 for(int i = tx;i;i >>= 1) 88 for(int j = ty;j;j >>= 1) 89 { 90 if(i == tx && j == ty)continue; 91 if(j == ty) 92 { 93 stx[i].sty[j].Min = min(stx[i<<1].sty[j].Min,stx[(i<<1)|1].sty[j].Min); 94 stx[i].sty[j].Max = max(stx[i<<1].sty[j].Max,stx[(i<<1)|1].sty[j].Max); 95 } 96 else 97 { 98 stx[i].sty[j].Min = min(stx[i].sty[j<<1].Min,stx[i].sty[(j<<1)|1].Min); 99 stx[i].sty[j].Max = max(stx[i].sty[j<<1].Max,stx[i].sty[(j<<1)|1].Max); 100 } 101 } 102 } 103 int queryMin(int i,int x1,int x2,int y1,int y2) 104 { 105 if(stx[i].l == x1 && stx[i].r == x2) 106 return stx[i].queryMin(1,y1,y2); 107 int mid = (stx[i].l + stx[i].r)/2; 108 if(x2 <= mid)return queryMin(i<<1,x1,x2,y1,y2); 109 else if(x1 > mid)return queryMin((i<<1)|1,x1,x2,y1,y2); 110 else return min(queryMin(i<<1,x1,mid,y1,y2),queryMin((i<<1)|1,mid+1,x2,y1,y2)); 111 } 112 int queryMax(int i,int x1,int x2,int y1,int y2) 113 { 114 if(stx[i].l == x1 && stx[i].r == x2) 115 return stx[i].queryMax(1,y1,y2); 116 int mid = (stx[i].l + stx[i].r)/2; 117 if(x2 <= mid)return queryMax(i<<1,x1,x2,y1,y2); 118 else if(x1 > mid)return queryMax((i<<1)|1,x1,x2,y1,y2); 119 else return max(queryMax(i<<1,x1,mid,y1,y2),queryMax((i<<1)|1,mid+1,x2,y1,y2)); 120 } 121 122 123 int main() 124 { 125 //freopen("in.txt","r",stdin); 126 //freopen("out.txt","w",stdout); 127 int T; 128 scanf("%d",&T); 129 int iCase = 0; 130 while(T--) 131 { 132 iCase++; 133 printf("Case #%d: ",iCase); 134 scanf("%d",&n); 135 build(1,1,n); 136 for(int i = 1;i <= n;i++) 137 for(int j = 1;j <= n;j++) 138 { 139 int a; 140 scanf("%d",&a); 141 Modify(i,j,a); 142 } 143 int q; 144 int x,y,L; 145 scanf("%d",&q); 146 while(q--) 147 { 148 scanf("%d%d%d",&x,&y,&L); 149 int x1 = max(x - L/2,1); 150 int x2 = min(x + L/2,n); 151 int y1 = max(y - L/2,1); 152 int y2 = min(y + L/2,n); 153 int Max = queryMax(1,x1,x2,y1,y2); 154 int Min = queryMin(1,x1,x2,y1,y2); 155 int t = (Max+Min)/2; 156 printf("%d ",t); 157 Modify(x,y,t); 158 } 159 } 160 return 0; 161 }