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  • Solution -「Code+#2」「洛谷 P4033」白金元首与独舞

    (mathcal{Description})

      link.
      给定一个 (n imes m) 的网格图,一些格子指定了走出该格的方向(上下左右),而有 (k) 格可以任意指定走出方向。求指定的方案数,使得从任意格子都可以走出网格图。
      (n,mle200;kle300)

    (mathcal{Solution})

      令“走出边界”为走到一个特殊点,建图,其中未指定方向的点向四周连边,相当于求以特殊点为根的内向树个数,跑矩阵树定理即可。复杂度 (mathcal O(n^3m^3))
      考虑优化,生成树个数实质上只与不定向的点有关,所以直接预处理出每个不定向点向上/下/左/右走到的第一个不定向点,向其连边,再跑矩阵树。复杂度 (mathcal O(k^3))

    (mathcal{Code})

    #include <cstdio>
    #include <cstring>
    #include <iostream>
    
    const int MOD = 1e9 + 7, MAXN = 200, MAXP = 300;
    int T, N, M, K[MAXP + 5][MAXP + 5], col[MAXN + 5][MAXN + 5], unk[MAXN + 5][MAXN + 5], cnt;
    char gar[MAXN + 5][MAXN + 5];
    
    inline int qkpow ( int a, int b, const int p = MOD ) {
    	int ret = 1;
    	for ( ; b; a = 1ll * a * a % p, b >>= 1 ) ret = 1ll * ret * ( b & 1 ? a : 1 ) % p;
    	return ret;
    }
    
    inline int det ( const int n ) {
    	int ret = 1, swp = 1;
    	for ( int i = 2; i <= n; ++ i ) {
    		for ( int j = i; j <= n; ++ j ) {
    			if ( K[j][i] ) {
    				if ( i ^ j ) std::swap ( K[i], K[j] ), swp *= -1;
    				break;
    			}
    		}
    		if ( ! ( ret = 1ll * ret * K[i][i] % MOD ) ) return 0;
    		for ( int j = i + 1, inv = qkpow ( K[i][i], MOD - 2 ); j <= n; ++ j ) {
    			int d = 1ll * K[j][i] * inv % MOD;
    			for ( int k = i; k <= n; ++ k ) K[j][k] = ( K[j][k] - 1ll * K[i][k] * d % MOD + MOD ) % MOD;
    		}
    	}
    	return ( ret * swp + MOD ) % MOD;
    }
    
    inline bool findLoop ( const int x, const int y, const int cur ) {
    	if ( x < 1 || x > N || y < 1 || y > M || gar[x][y] == '.' ) return false;
    	if ( col[x][y] == cur ) return true;
    	if ( col[x][y] ) return false;
    	col[x][y] = cur;
    	if ( gar[x][y] == 'L' ) return findLoop ( x, y - 1, cur );
    	if ( gar[x][y] == 'R' ) return findLoop ( x, y + 1, cur );
    	if ( gar[x][y] == 'U' ) return findLoop ( x - 1, y, cur );
    	if ( gar[x][y] == 'D' ) return findLoop ( x + 1, y, cur );
    	return false;
    }
    
    inline int findUnknown ( const int x, const int y ) {
    	if ( x < 1 || x > N || y < 1 || y > M ) return 1;
    	if ( unk[x][y] ) return unk[x][y];
    	int& ret = unk[x][y];
    	if ( gar[x][y] == 'L' ) ret = findUnknown ( x, y - 1 );
    	if ( gar[x][y] == 'R' ) ret = findUnknown ( x, y + 1 );
    	if ( gar[x][y] == 'U' ) ret = findUnknown ( x - 1, y );
    	if ( gar[x][y] == 'D' ) ret = findUnknown ( x + 1, y );
    	return ret;
    }
    
    inline void add ( const int s, int t ) {
    	if ( ! t ) t = 1;
    	++ K[s][s], -- K[s][t];
    	if ( K[s][t] < 0 ) K[s][t] += MOD;
    }
    
    int main () {
    	for ( scanf ( "%d", &T ); T --; ) {
    		memset ( K, 0, sizeof K );
    		memset ( col, 0, sizeof col );
    		memset ( unk, 0, sizeof unk );
    		scanf ( "%d %d", &N, &M ), cnt = 1;
    		for ( int i = 1; i <= N; ++ i ) {
    			scanf ( "%s", gar[i] + 1 );
    			for ( int j = 1; j <= M; ++ j ) {
    				if ( gar[i][j] == '.' ) {
    					unk[i][j] = ++ cnt;
    				}
    			}
    		}
    		bool loop = false;
    		for ( int i = 1, cur = 1; i <= N && ! loop; ++ i ) {
    			for ( int j = 1; j <= M && ! loop; ++ j ) {
    				loop |= findLoop ( i, j, cur ++ );
    			}
    		}
    		if ( loop ) { puts ( "0" ); continue; }
    		for ( int i = 1; i <= N; ++ i ) {
    			for ( int j = 1; j <= M; ++ j ) {
    				findUnknown ( i, j );
    			}
    		}
    		for ( int i = 1; i <= N; ++ i ) {
    			for ( int j = 1; j <= M; ++ j ) {
    				if ( gar[i][j] == '.' ) {
    					add ( unk[i][j], unk[i][j - 1] );
    					add ( unk[i][j], unk[i][j + 1] );
    					add ( unk[i][j], unk[i - 1][j] );
    					add ( unk[i][j], unk[i + 1][j] );
    				}
    			}
    		}
    		printf ( "%d
    ", det ( cnt ) );
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/rainybunny/p/13224064.html
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