暴力/思维 HDOJ 5386 Cover

题目传送门

 /*
     题意：给出刷墙的所有的方法，求一种顺序，使得原矩阵刷成目标矩阵
     暴力：(题解)我们只要每次找一行或一列颜色除了0都相同的，然后如果有对应的操作，就把这行这列都赋值成0即可
 */
 /************************************************
 * Author        :Running_Time
 * Created Time  :2015-8-13 21:25:07
 * File Name     :G.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 1e2 + ;
 const int MAXM = 5e2 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 int a[MAXN][MAXN];
 int b[MAXM], c[MAXM];
 char op[MAXM];
 int ans[MAXM];

 int main(void)    {     //HDOJ 5386 Cover
     int T;  scanf ("%d", &T);
     while (T--) {
         int n, m;   scanf ("%d%d", &n, &m);
         for (int i=; i<=n; ++i)    {
             for (int j=; j<=n; ++j)    {
                 scanf ("%d", &a[i][j]);
             }
         }
         for (int i=; i<=n; ++i)    {
             for (int j=; j<=n; ++j)    {
                 scanf ("%d", &a[i][j]);
             }
         }
         for (int i=; i<=m; ++i)    {
             char ch;
             for (ch=getchar (); ch!='H'&&ch!='L'; ch=getchar ()) ;
             op[i] = ch; scanf ("%d%d", &b[i], &c[i]);
         }

         int t = ;
         while (t < m)   {
             for (int i=; i<=m; ++i)    {
                 if (!b[i])  continue;
                 if (op[i] == 'H')   {
                     int x = b[i];   bool flag = true;
                     for (int j=; j<=n; ++j)    {
                         if (a[x][j] && a[x][j] != c[i])     {
                             flag = false; break;
                         }
                     }
                     if (flag)   {
                         ans[++t] = i;   b[i] = ;
                         for (int j=; j<=n; ++j)    a[x][j] = ;
                     }
                 }
                 else    {
                     int y = b[i]; bool flag = true;
                     for (int j=; j<=n; ++j)    {
                         if (a[j][y] && a[j][y] != c[i]) {
                             flag = false;   break;
                         }
                     }
                     if (flag)   {
                         ans[++t] = i;   b[i] = ;
                         for (int j=; j<=n; ++j)    a[j][y] = ;
                     }
                 }
             }
         }

         for (int i=m; i>=; --i)    printf ("%d%c", ans[i], (i == ) ? '\n' : ' ');
     }

     return ;
 }