sgu 125

AC之后看大网上别人的基本是dfs、暴力搜索

因为我的算法不是这样，贴出来看看

sgu结果：  Accepted  31 ms   835 kb

 

英文很烂，注释看不懂可email

 

/** * @file * @brief my c solution for problem 125 on SGU OJ * @author ck<[e-mail]> * @date 2013-06-19 * @version  *  * @section DESCRIPTION * * There must exist zero in B, if not, the problem has no answer. * The A value in the zero position must be bigger than others. * Put the biggest value to A at the zero position, update B around * the zero positon. Continue this untill A is all filled. * Here is the algrithm: *  1, init *      biggest_value = 9; *      A = 0;    // zero in A means that it is not filled, we can always *                   fill it with a negtive number. *      zero[] = 0; // mark zero positons. *                  // zero position means a positon where A and B                     // are both zero *                  // 0: not a zero positon, *                  // 1: a zero positon, its neighbours ar not checked *                  // 2: a zero positon, its neighbours are checked *  2, find a positon where A and B are both zero *      if exist *           mark the position with 1, goto 3 *      else if A is all filled  *           success! output A  *      else  *           NO_ANS * 3, extend zero positon in the same row or column, extend in four  *    directions -- up, down, right, left *        1) find a zero position mark with 1. *        2) check if its four neighbours are zero positon *           if it is, and its zero mark is zero, *                mark the positon with 1; *        3) mark this positon with 2 *        4) goto 1), untill you cannot find zero positon mark with 1. * 4, put biggest_value to A at zero positons * 5, biggest_value = biggest_value - 1; * 6, update B with zero positons *      foreach zero positon *         if its neighbour value in A is zero *              sub B value with 1 *              if B value < 0 *                 NO_ANS * 7, goto 2 */#include <stdio.h>#include <string.h>#define NO_ANS printf("NO SOLUTION\n");return 0;#define SZ 3        /**< max size of input */int n;   int A[SZ][SZ] = {{0}};int B[SZ][SZ] = {{0}};int i, j, nn, row, col;int element = SZ*SZ;  /**< the number need to put to A */int zero[SZ*SZ];    /**< zeros positon row*n+col */int main() {  // read input  scanf("%d", &n);  nn = n * n;  for (i=0; i<n; ++i) {    for (j=0; j<n; ++j) {      scanf("%d", &B[i][j]);    }  }  while (element > 0) {    // find zeros    row = 0;   //indicate if there is zero    memset(zero, 0, sizeof(zero));    for (i=0; i<n; ++i) {      for (j=0; j<n; ++j) {if (B[i][j] == 0 && A[i][j] == 0) {  row = 1;  zero[i*n+j] = 1;  break;}      }      if (row > 0) {break;      }    }    // cannot find zero    if (row <= 0) {      for (i=0; i<n; ++i) {for (j=0; j<n; ++j) {  if (A[i][j] <= 0) {    NO_ANS;  }}      }      break;    }    // can find zero, extend    j = 1;  // indicate if zero can be extened    while (j) {      j = 0;      for (i=0; i<nn; ++i) {if (zero[i] == 1) {  row = i / n;  col = i % n;  j = row - 1;  if (j>=0 && A[j][col]==0 && B[j][col]==0 && zero[j*n+col]==0) {    zero[j*n+col] = 1;  }  j = row + 1;  if (j<n && A[j][col]==0 && B[j][col]==0 && zero[j*n+col]==0) {    zero[j*n+col] = 1;  }  j = col - 1;  if (j>=0 && A[row][j]==0 && B[row][j]==0 && zero[row*n+j]==0) {    zero[row*n+j] = 1;  }  j = col + 1;  if (j<n && A[row][j]==0 && B[row][j]==0 && zero[row*n+j]==0) {    zero[row*n+j] = 1;  }  zero[i] = 2;  j = 1;}      }    }    // put element to A    for (i=0; i<nn; ++i) {      if (zero[i]) {row = i / n;col = i % n;A[row][col] = element;      }    }    --element;    // update B    for (i=0; i<nn; ++i) {      if (zero[i]) {row = i / n;col = i % n;      j = row - 1;if (j>=0 && A[j][col]<=0) {  --B[j][col];  if (B[j][col] < 0) {    NO_ANS;  }}j = row + 1;if (j<n && A[j][col]<=0) {  --B[j][col];  if (B[j][col] < 0) {    NO_ANS;  }}      j = col - 1;if (j>=0 && A[row][j]<=0) {  --B[row][j];  if (B[row][j] < 0) {    NO_ANS;  }}j = col + 1;if (j<n && A[row][j]<=0) {  --B[row][j];  if (B[row][j] < 0) {    NO_ANS;  }}      }    }  }  // output ans  for (i=0; i<n; ++i) {    for (j=0; j<n; ++j) {      printf("%d ", A[i][j]);    }    putchar('\n');  }  return 0;}