【POJ 2411】 Mondriaan's Dream

【题目链接】

          点击打开链接

【算法】

         很明显，我们可以用状态压缩动态规划解决此题

         f[n][m]表示n-1行已经放满，第n行状态为m的合法的方案数

         状态转移方程很好推

         注意这题时限较紧，注意加一些小优化

 【代码】

            

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 12

long long n,m;
long long f[MAXN][1<<MAXN],ans[MAXN][MAXN];

template <typename T> inline void read(T &x) {
        long long f = 1; x = 0;
        char c = getchar();
        for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
        for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
        x *= f;
}
template <typename T> inline void write(T x) {
        if (x < 0) { putchar('-'); x = -x; }
        if (x > 9) write(x/10);
        putchar(x%10+'0');    
}
template <typename T> inline void writeln(T x) {
        write(x);
        puts("");    
}
inline bool ok(long long s) {
        long long i = 0;
        while (i < m) {
                if (s & (1 << i)) {
                        if (!(s & (1 << (i + 1))))
                                return false;
                        i += 2;
                        continue;
                }
                i++;
        }    
        return true;
}
inline bool check(long long now,long long last) {
        long long i = 0;
        while (i < m) {
                if (!(last & (1 << i))) {
                        if (!(now & (1 << i))) return false;
                        i++;
                        continue;
                } else {
                        if ((now & (1 << i)) && (now & (1 << (i + 1))) && (last & (1 << (i + 1)))) {
                                i += 2;
                                continue;
                        }
                        if (!(now & (1 << i))) {
                                i++;
                                continue;
                        }
                        return false;
                }
        }
        return true;
}
inline void solve(long long n,long long m) {
        long long i,j,k,
            MASK = (1 << m) - 1;        
        for (i = 0; i <= MASK; i++) if (ok(i)) f[1][i] = 1; else f[1][i] = 0;
        for (i = 2; i <= n; i++) {
                for (j = 0; j <= MASK; j++) {
                        f[i][j] = 0;
                        for (k = 0; k <= MASK; k++) {
                                if (check(j,k))
                                        f[i][j] += f[i-1][k];
                        }
                }
        }
        ans[n][m] = ans[m][n] = f[n][MASK];
        writeln(f[n][MASK]);
}

int main() {

        memset(ans,255,sizeof(ans));
        while (true) {
                read(n); read(m);
                if (!n && !m) break;
                if (n & 1 && m & 1) ans[n][m] = ans[m][n] = 0;
                if (ans[n][m] != -1) {
                        writeln(ans[n][m]);
                        continue;
                }
                if (n < m) swap(n,m);
                solve(n,m);    
        }
    
        return 0;    
}