Corn Fields (like pressure DP)

Corn Fields

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 23437		Accepted: 12275

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

Line 1: Two space-separated integers: M and N
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

Line 1: One integer: the number of ways that FJ can choose the squares modulo 100,000,000.

Sample Input

2 3
1 1 1
0 1 0

Sample Output

Hint

Number the squares as follows:

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.

Problem-solving ideas: Title row mean no conflict and no conflict columns can only be placed at position 1 (scheme 1 hold) the total specific look at the code output scheme

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 
 6 using namespace std;
 7 int n,m;
 8 const int maxn=13;
 9 const int mod=1e8;
10 int arr[15][15];
11 int state[1<<maxn];
12 int ee[maxn];   //每一行的状态
13 intDP [MAXN] [ . 1 << MAXN];   // number of i-th row j-th states adopted solution 
14  
15  
16  int main () {
 . 17      iOS :: sync_with_stdio ( to false );
 18 is      DP [ 0 ] [ 0 ] = 1 ;   // all not fill to 1 
. 19      CIN >> >> n- m;
 20 is      for ( int I = 1 ; I <= n-; I ++ ) {
 21 is          for ( int J = 1 ; J <= m; J ++ ) {
 22 is              CIN >> ARR [I] [J];
 23 is          }
24      }
 25      int len = . 1 << m;
 26 is      for ( int I = 0 ; I <len; I ++) {    // each state that it will not conflict line 
27          State [I] = ((I & (I < < . 1 )) == 0 && (I & (I >> . 1 )) == 0 ); // . 1 means no conflict, the conflict represents 0 
28      } 
 29      for ( int I = . 1 ; I <= n-; I ++ ) {
 30          for ( int J = . 1 ; J <= m; J ++ ) {
 31 is              EE [I] = (EE [I] <<. 1 ) | ARR [I] [J];    // status of each line 
32          }
 33 is      }
 34 is      for ( int I = . 1 ; I <= n-; I ++ ) {
 35          for ( int J = 0 ; J <len; J ++ ) {
 36              IF (state [J] && (J & EE [i]) == J) {    // first, no conflict followed state to state i rows 
37 [                  for ( int K = 0 ; K <len; K ++ ) {
 38 is                      IF ((J & K) == 0 ) {    // this line and there is no conflict on the column line 
39                          DP [I] [J] = (DP [I] [J] + DP [I-. 1 ] [K])% MOD;    // Do not forget to take the remainder 
40                      }
 41 is                  }
 42 is              }
 43 is          }
 44 is      }
 45      int RES = 0 ;
 46 is      for ( int I = 0 ; I <len; I ++ ) {
 47          RES + = DP [n-] [I]; 
 48          RES% = MOD;    // Do not forget to take the remainder 
49      }
 50      COUT RES << << endl;
 51 is      return  0 ;
 52 is }

View Code

Corn Fields (like pressure DP)

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