| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
|
D |
Even Parity Input: Standard Input Output: Standard Output |
We have a grid of size N x N. Each cell of the grid initially
contains a zero(0) or a one(1).
The parity of a cell
is the number of 1s surrounding that cell. A cell is surrounded by at most 4
cells (top, bottom, left, right).
Suppose we have a grid of size 4 x 4:
|
1 |
0 |
1 |
0 |
The parity of each cell would be |
1 |
3 |
1 |
2 |
|
1 |
1 |
1 |
1 |
2 |
3 |
3 |
1 | |
|
0 |
1 |
0 |
0 |
2 |
1 |
2 |
1 | |
|
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
For this problem, you have to change some of the 0s to 1s so that the parity of every cell becomes even. We are interested in the minimum number of transformations of 0 to 1 that is needed to achieve the desired requirement.
The first line of input is an integer T (T<30) that indicates the number of test cases. Each case starts with a positive integer N(1≤N≤15). Each of the next N lines contain N integers (0/1) each. The integers are separated by a single space character.
For each case, output the case number followed by the minimum number of transformations required. If it‘s impossible to achieve the desired result, then output -1 instead.
3 3 0 0 0 0 0 0 0 0 0 3 0 0 0 1 0 0 0 0 0 3 1 1 1 1 1 1 0 0 0
|
Case 1: 0 |
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 #include<time.h> 5 #include<iostream> 6 #include<algorithm> 7 using namespace std; 8 int a[20][20],n,b[20][20]; 9 long long ab; 10 int fun1(int x,int y) 11 { 12 return b[x][y+1]^b[x][y-1]^b[x-1][y]; 13 } 14 int fun(int x) 15 { 16 if((ab&x)!=ab)return 11111110; 17 memset(b,0,sizeof(b)); 18 int i,j,ans=0; 19 for(i=n; i>=1; i--) 20 { 21 b[1][i]=x&1; 22 if(b[1][i]^a[1][i]) 23 { 24 if(b[1][i])ans++; 25 } 26 x>>=1; 27 } 28 for(i=2; i<=n; i++) 29 { 30 for(j=1; j<=n; j++) 31 { 32 b[i][j]=fun1(i-1,j); 33 if(b[i][j]^a[i][j]) 34 { 35 if(b[i][j]) 36 ans++; 37 else return 11111110; 38 } 39 } 40 } 41 return ans; 42 } 43 int main() 44 { 45 int T,i,j,k,size,ans; 46 scanf("%d",&T); 47 for(i=1; i<=T; i++) 48 { 49 ans=11111110; 50 memset(a,0,sizeof(a)); 51 scanf("%d",&n); 52 for(j=1; j<=n; j++) 53 for(k=1; k<=n; k++) 54 scanf("%d",&a[j][k]); 55 size=(1<<n); 56 ab=0; 57 for(j=1; j<=n; j++) 58 { 59 ab=(ab<<1)+a[1][j]; 60 } 61 for(j=0; j<size; j++) 62 { 63 int y=fun(j); 64 ans=ans>y?y:ans; 65 } 66 if(ans==11111110)ans=-1; 67 printf("Case %d: %d\n",i,ans); 68 } 69 }
Even Parity uva11464 模拟,布布扣,bubuko.com
原文：http://www.cnblogs.com/ERKE/p/3683810.html
