ZOJ 1063 Space Station Shielding

練習floodfill很好的題。第一次考慮從內部有模塊的點開始floodfill，發現無法識別內部空倉和外部空倉。反向思考，從外部空倉進入模塊，得解。

 

#include <cstdio> #include <string> int b[60][60][60]; bool used[216000]; int q[216000]; int N, M, K, L; int dir[6][3] = { { 1, 0, 0 }, { -1, 0, 0 }, { 0, 1, 0 }, { 0, -1, 0 }, { 0, 0, 1 }, { 0, 0, -1 } }; int getx ( int n, int m, int k ) { return ( k * M + m ) * N + n; } void setx ( int x, int &n, int &m, int &k ) { k = x / ( N * M ); x = x % ( N * M ); m = x / N; n = x % N; } int main () { //freopen ( "in.txt", "r", stdin ); while ( scanf ( "%d%d%d%d", &N, &M, &K, &L ) && N ) { memset ( b, 0, sizeof ( b ) ); int i, x, n, m, k; for ( i = 0; i < L; i ++ ) { scanf ( "%d", &x ); setx ( x, n, m, k ); b[n][m][k] = 1; } memset ( used, 0, sizeof ( used ) ); int op, ed; // 對外層平面的所有0值點進行floodfill，進入1值點的即計數 int count = 0; for ( n = 0; n < N; n ++ ) { for ( m = 0; m < M; m ++ ) { for ( k = 0; k < K; k ++ ) { if ( n != 0 && n != N - 1 && m != 0 && m != M - 1 && k != 0 && k != K - 1 ) continue; if ( b[n][m][k] == 0 ) { b[n][m][k] = 2; int x = getx ( n, m, k ); q[0] = x; for ( op = -1, ed = 0; op ++ < ed; ) { int xx = q[op]; int kk, mm, nn; setx ( xx, nn, mm, kk ); int l; for ( l = 0; l < 6; l ++ ) { int nnn = nn + dir[l][0], mmm = mm + dir[l][1], kkk = kk + dir[l][2]; if ( nnn < 0 || nnn >= N || mmm < 0 || mmm >= M || kkk < 0 || kkk >= K ) continue; if ( b[nnn][mmm][kkk] == 1 ) count ++; if ( b[nnn][mmm][kkk] == 0 ) { b[nnn][mmm][kkk] = 2; int xxx = getx ( nnn, mmm, kkk ); q[++ ed] = xxx; } } } } } } } for ( n = 0; n < N; n ++ ) { for ( m = 0; m < M; m ++ ) { if ( b[n][m][0] == 1 ) count ++; if ( b[n][m][K - 1] == 1 ) count ++; } } for ( k = 0; k < K; k ++ ) { for ( m = 0; m < M; m ++ ) { if ( b[0][m][k] == 1 ) count ++; if ( b[N - 1][m][k] == 1 ) count ++; } } for ( n = 0; n < N; n ++ ) { for ( k = 0; k < K; k ++ ) { if ( b[n][0][k] == 1 ) count ++; if ( b[n][M - 1][k] == 1 ) count ++; } } printf ( "The number of faces needing shielding is %d./n", count ); } return 0; }