ZOJ 2563 Long Dominoes（状压DP）题解

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 11 + 5;
const int M = maxn * 30;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const int MOD = 1e4 + 7;
ll dp[31][1 << maxn];
int fac[maxn];
int vis[maxn];
int n, m;
int prest;
//0:都是空的 1:上一个非空 2:都非空
int getSt(){
    int ret = 0;
    for(int i = m - 1; i >= 0; i--)
        ret = ret * 3 + vis[i];
    return ret;
}
void dfs(int i, int j){
    if(j >= m){
        int st = getSt();
        dp[i][st] += dp[i - 1][prest];
        return;
    }
    int p = prest / fac[j] % 3;
    if(p == 0){
        vis[j] = 2;
        dfs(i, j + 1);
    }
    else if(p == 1){
        vis[j] = 0;
        dfs(i, j + 1);
    }
    else{
        if(j >= 2 && vis[j - 1] == 1 && vis[j - 2] == 1){
            vis[j] = vis[j - 1] = vis[j - 2] = 2;
            dfs(i, j + 1);
            vis[j] = vis[j - 1] = vis[j - 2] = 1;
        }
        vis[j] = 1;
        dfs(i, j + 1);
    }
}
int main(){
    fac[0] = 1;
    for(int i = 1; i <= 10; i++) fac[i] = fac[i - 1] * 3;
    while(~scanf("%d%d", &m, &n) && n + m){
        memset(dp, 0, sizeof(dp));
        dp[0][fac[m] - 1] = 1;
        for(int i = 1; i <= n; i++){
            for(int j = 0; j < fac[m]; j++){
                if(dp[i - 1][j] == 0) continue;
                memset(vis, 0, sizeof(vis));
                prest = j;
                dfs(i, 0);
            }
        }
        printf("%lld\n", dp[n][fac[m] - 1]);
    }
    return 0;
}