| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 9806 | Accepted: 5185 |
Description
Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares are infertile and can't be planted. Canny FJ knows that the cows dislike eating close to each other, so when choosing which squares to plant, he avoids choosing squares that are adjacent; no two chosen squares share an edge. He has not yet made the final choice as to which squares to plant.
Being a very open-minded man, Farmer John wants to consider all possible options for how to choose the squares for planting. He is so open-minded that he considers choosing no squares as a valid option! Please help Farmer John determine the number of ways he can choose the squares to plant.
Input
Lines 2.. M+1: Line i+1 describes row i of the pasture with N space-separated integers indicating whether a square is fertile (1 for fertile, 0 for infertile)
Output
Sample Input
2 3 1 1 1 0 1 0
Sample Output
9
Hint
1 2 3
4
There are four ways to plant only on one squares (1, 2, 3, or 4), three ways to plant on two squares (13, 14, or 34), 1 way to plant on three squares (134), and one way to plant on no squares. 4+3+1+1=9.
Source
1 #include <stdio.h> 2 #define MOD 100000000 3 int row[13], rec[377], dp[13][377]; 4 int main() 5 { 6 int x=1<<12, k=0; 7 for(int i=0; i<x; i++) //calculate all valid states 8 if(!(i&(i<<1))) //is a valid row state 9 rec[k++]=i; 10 rec[k]=x; 11 12 int M, N, t; 13 scanf("%d%d", &M, &N); 14 for(int i=0; i<M; i++) 15 for(int j=0; j<N; j++) 16 scanf("%d", &t), //t = Matrix[i][j] 17 row[i]=(row[i]<<1)|!t; //reverse row state 18 19 x=1<<N; 20 for(int i=0; rec[i]<x; i++) //process first row 21 if(!(row[0]&rec[i])) 22 dp[0][i]=1; 23 for(int r=1; r<M; r++) //for each row 24 for(int i=0; rec[i]<x; i++) //for each valid state for last row 25 if(!(row[r-1]&rec[i])) 26 for(int j=0; rec[j]<x; j++) //for each valid state for this row 27 if(!(row[r]&rec[j])) 28 if(!(rec[i]&rec[j])) //the two states are not conflict 29 dp[r][j]=(dp[r][j]+dp[r-1][i])%MOD; 30 31 int r=M-1; 32 for(int i=1; rec[i]<x; i++) //process last row 33 dp[r][0]=(dp[r][0]+dp[r][i])%MOD; 34 printf("%d\n", dp[r][0]); 35 36 return 0; 37 }