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POJ 3254 Corn Fields

//狀態壓縮
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#include <cstdio>
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int N, M;
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struct ITEM
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{
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int m_arr[1024];
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int m_size;
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}item[13];
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int dp[13][1024];
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//生成每行的所有合法狀態
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//參數：pos 代表行 illegal代表不合法的狀態(源行為0的狀態)
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void GetState(const int& pos, const int& illegal)
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{
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int state, n;
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n = 0;
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for ( state = 0; state < (1 << N); ++state ) //從0開始枚舉到(1 << N): N = 3時, 000-->111 ((1 << N)8 - 1)
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{
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if ( (state << 1) & state || (state >> 1) & state ) //去除相鄰都為1的狀態
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continue;
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if ( state & illegal ) //去除原為0的狀態
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continue;
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item[pos].m_arr[n++] = state;
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}
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item[pos].m_size = n;
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}
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void Solve()
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{
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int i, j, k;
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//初始化：首行的每個合法狀態必為1
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for ( i = 0; i < item[0].m_size; ++i )
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dp[0][i] = 1;
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//狀態方程：dp[i][j] = SUM{dp[i - 1][k]}
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for ( i = 1; i < M; ++i )
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{
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for ( j = 0; j < item[i].m_size; ++j )
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{
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dp[i][j] = 0;
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for ( k = 0; k < item[i - 1].m_size; ++k )
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{
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//與上一行沖突
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if ( item[i].m_arr[j] & item[i - 1].m_arr[k] )
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continue;
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dp[i][j] += dp[i - 1][k];
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}
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}
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}
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int result = 0;
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for ( i = 0; i < item[M - 1].m_size; ++i )
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{
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result += dp[M - 1][i];
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result %= 100000000;
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}
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printf("%d\n", result);
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}
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int main()
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{
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// freopen("1.txt", "r", stdin);
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int i, j, state, illegal;
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scanf("%d %d", &M, &N);
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for ( i = 0; i < M; ++i )
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{
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illegal = 0;
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for ( j = 0; j < N; ++j )
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{
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scanf("%d", &state);
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//取反狀態
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illegal = (illegal << 1) + 1 - state;
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}
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GetState( i, illegal );
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}
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Solve();
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return 0;
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}

posted on 2009-05-02 10:51 巫閱讀(351) 評論(0) 編輯收藏引用所屬分類: DP

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