状态压缩 dp
dp[i][j] 为第 i 行状态为 j 的总数
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 2005
#define INF 0x3f3f3f3f
#define inf 10000000
#define MOD 100000000
#define ULL unsigned long long
#define LL long long
using namespace std;
int dp[15][1 << 13], mapp[1 << 13], sta[1 << 13];
int n, m, num;
void init() {
memset(dp, 0, sizeof(dp));
memset(mapp, 0, sizeof(mapp));
num = 0;
for(int i = 0; i < (1<<m); ++ i) {
int d = i&(i << 1);
if(d == 0) sta[num++] = i;
}
// printf("num : %d\n", num);
}
int main()
{
while(scanf("%d%d", &n, &m) == 2) {
init();
// puts("*********************");
// for(int i = 0; i < num; ++ i) {
// printf("%d ", sta[i]);
// }
// puts("**********************");
for(int i = 0; i < n; ++ i) {
for(int j = 0; j < m; ++ j) {
int a;
scanf("%d", &a);
if(a == 0) mapp[i] = mapp[i]|(1 << j);
}
}
for(int i = 0; i < num; ++ i) {
if((mapp[0] & sta[i]) == 0) {
dp[0][i] = 1;
}
}
for(int i = 1; i < n; ++ i) {
for(int j = 0; j < num; ++ j) {
if(mapp[i-1] & sta[j]) continue;
for(int q = 0; q < num; ++ q) {
if(mapp[i]&sta[q] || sta[j]&sta[q]) continue;
dp[i][q] += dp[i-1][j];
dp[i][q] %= MOD;
}
}
}
int ans = 0;
for(int i = 0; i < num; ++ i) {
ans += dp[n-1][i];
ans %= MOD;
}
printf("%d\n", ans);
}
return 0;
}
原文:http://www.cnblogs.com/avema/p/3862824.html