题目描述
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.
农场主John新买了一块长方形的新牧场，这块牧场被划分成M行N列(1 ≤ M ≤ 12; 1 ≤ N ≤ 12)，每一格都是一块正方形的土地。John打算在牧场上的某几格里种上美味的草，供他的奶牛们享用。
遗憾的是，有些土地相当贫瘠，不能用来种草。并且，奶牛们喜欢独占一块草地的感觉，于是John不会选择两块相邻的土地，也就是说，没有哪两块草地有公共边。
John想知道，如果不考虑草地的总块数，那么，一共有多少种种植方案可供他选择？（当然，把新牧场完全荒废也是一种方案）
输入格式
第一行：两个整数M和N，用空格隔开。
第2到第M+1行：每行包含N个用空格隔开的整数，描述了每块土地的状态。第i+1行描述了第i行的土地，所有整数均为0或1，是1的话，表示这块土地足够肥沃，0则表示这块土地不适合种草。
输出格式
一个整数，即牧场分配总方案数除以100,000,000的余数。
输入输出样例
输入 #1复制
2 3
1 1 1
0 1 0
输出 #1复制
9
看到数据范围想到状压，设dp[i, j]为第i行状态为j的方案数。首先，题目限制了只能在特定的土地上种草，因此所有可行状态的并S需要通过位运算求出来。然后对S进行子集枚举，设枚举到的状态为p。但是题目还限定了另外一个条件，就是相邻的土地不能都种草。因此如果p中有相邻的1的话则dp[i, p] = 0。那么怎么判定p能否从前一行转移过来呢？可以用(1 << m) - 1与p进行异或得到前一行的可行状态_p，然后再对_p进行子集枚举。注意这里是用(1 << m) - 1与p异或而非S，道理很简单，看样例就能明白了～～
需要注意的是子集枚举不会枚举到0，而本题中不种草的话对于答案也有贡献，因此需要特殊处理。
#include <iostream>
#include <cstring>
#define mod 100000000
using namespace std;
int n, m;
int mp[15][15];
long long dp[15][(1 << 12)], sum[15];
bool adjacent(int x) {
//判断x中是否有相邻的1
for(int i = 1; i < 32; i++) {
if(((x >> i) & 1) && ((x >> (i - 1)) & 1)) return 1;
}
return 0;
}
int main() {
memset(dp, 0, sizeof(dp));
memset(sum, 0, sizeof(sum));
cin >> n >> m;
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= m; j++) {
cin >> mp[i][j];
}
}
for(int i = 0; i < (1 << 12); i++) dp[0][i] = 1ll;
sum[0] = 1;
for(int i = 1; i <= n; i++) {
int S = 0;
for(int j = 0; j < m; j++) {
S += (mp[i][j + 1] << j);
}
for(int p = S; p; p = (p - 1) & S) {//子集枚举不会枚举到0
// to calc dp[i][p]
if(adjacent(p)) continue;
if(i == 1) {
dp[i][p] = 1ll;
sum[i] = (sum[i] + 1) % mod;
continue;
}
int _p = ((1 << m) - 1) ^ p;
for(int q = _p; q; q = (q - 1) & _p) {//子集枚举不会枚举到0
dp[i][p] = (dp[i - 1][q] + dp[i][p]) % mod;
}
dp[i][p] = (dp[i - 1][0] + dp[i][p]) % mod;
sum[i] = (sum[i] + dp[i][p]) % mod;
}
dp[i][0] = sum[i - 1];
sum[i] = (sum[i] + dp[i][0]) % mod;
}
cout << sum[n];
return 0;
}