暴力/思维 HDOJ 5386 Cover-白红宇

暴力/思维 HDOJ 5386 Cover

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发布时间：2019-06-24

本文共 2118 字，大约阅读时间需要 7 分钟。

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 
     
      12 #include 
      
       13 #include 
       
        14 #include 
        
         15 #include 
         
          16 #include 
          
           17 #include 
           
            18 #include 
            
             19 #include 
             
              20 #include 
              
               21 #include 
               
                22 #include 
                
                 23 #include 
                 24 #include 
                  
                   25 #include 
                   
                    26 #include 
                    
                     27 #include 
                     
                      28 using namespace std;29 30 #define lson l, mid, rt << 131 #define rson mid + 1, r, rt << 1 | 132 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 Cover43 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) ? '\n' : ' ');94 }95 96 return 0;97 }