暴力/思维 HDOJ 5386 Cover

 1 /*
 2     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
 3     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    
 4 */
 5 /************************************************
 6 * Author        :Running_Time
 7 * Created Time  :2015-8-13 21:25:07
 8 * File Name     :G.cpp
 9  ************************************************/
10 
11 #include <cstdio>
12 #include <algorithm>
13 #include <iostream>
14 #include <sstream>
15 #include <cstring>
16 #include <cmath>
17 #include <string>
18 #include <vector>
19 #include <queue>
20 #include <deque>
21 #include <stack>
22 #include <list>
23 #include <map>
24 #include <set>
25 #include <bitset>
26 #include <cstdlib>
27 #include <ctime>
28 using namespace std;
29 
30 #define lson l, mid, rt << 1
31 #define rson mid + 1, r, rt << 1 | 1
32 typedef long long ll;
33 const int MAXN = 1e2 + 10;
34 const int MAXM = 5e2 + 10;
35 const int INF = 0x3f3f3f3f;
36 const int MOD = 1e9 + 7;
37 int a[MAXN][MAXN];
38 int b[MAXM], c[MAXM];
39 char op[MAXM];
40 int ans[MAXM];
41 
42 int main(void)    {     //HDOJ 5386 Cover
43     int T;  scanf ("%d", &T);
44     while (T--) {
45         int n, m;   scanf ("%d%d", &n, &m);
46         for (int i=1; i<=n; ++i)    {
47             for (int j=1; j<=n; ++j)    {
48                 scanf ("%d", &a[i][j]);
49             }
50         }
51         for (int i=1; i<=n; ++i)    {
52             for (int j=1; j<=n; ++j)    {
53                 scanf ("%d", &a[i][j]);
54             }
55         }
56         for (int i=1; i<=m; ++i)    {
57             char ch;
58             for (ch=getchar (); ch!='H'&&ch!='L'; ch=getchar ()) ;
59             op[i] = ch; scanf ("%d%d", &b[i], &c[i]);
60         }
61         
62         int t = 0;
63         while (t < m)   {
64             for (int i=1; i<=m; ++i)    {
65                 if (!b[i])  continue;
66                 if (op[i] == 'H')   {
67                     int x = b[i];   bool flag = true;
68                     for (int j=1; j<=n; ++j)    {
69                         if (a[x][j] && a[x][j] != c[i])     {
70                             flag = false; break;
71                         }
72                     }
73                     if (flag)   {
74                         ans[++t] = i;   b[i] = 0;
75                         for (int j=1; j<=n; ++j)    a[x][j] = 0;
76                     }
77                 }
78                 else    {
79                     int y = b[i]; bool flag = true;
80                     for (int j=1; j<=n; ++j)    {
81                         if (a[j][y] && a[j][y] != c[i]) {
82                             flag = false;   break;
83                         }
84                     }
85                     if (flag)   {
86                         ans[++t] = i;   b[i] = 0;
87                         for (int j=1; j<=n; ++j)    a[j][y] = 0;
88                     }
89                 }
90             }
91         }
92 
93         for (int i=m; i>=1; --i)    printf ("%d%c", ans[i], (i == 1) ? '
' : ' ');
94     }
95 
96     return 0;
97 }
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