URAL 1957 Wrong Answer 暴力

Wrong Answer

思路：

1.先枚举4的全排列，即球赛的所有可能结果，一共4!=24种情况

2.对于每种情况，DFS 未确定的比赛 的结果，判断这种情况是否可达。

剪枝：

1.对于每种全排列，只需要找到一种满足这个全排列的情况即可，如果有一种情况满足，则不必再枚举其他可能。

2.因为在每种情况下，球队之间的排名已经确定了，所以在DFS的过程中，凡是不满足此排名的情况统统可以剪掉。

 

这样就可以少很多情况，即使在n=0的情况下也跑出了不错的效率，但是它Wrong Answer on test 17了。。。

求指点T^T

 

  1 #include <cstdio>

  2 #include <cstring>

  3 #include <cstdlib>

  4 #include <algorithm>

  5 

  6 using namespace std;

  7 

  8 const int MAXN = 10010;

  9 const int permuta[30] = { 1234, 1243, 1324, 1342, 1423, 1432,

 10                           2134, 2143, 2314, 2341, 2413, 2431,

 11                           3124, 3142, 3214, 3241, 3412, 3421,

 12                           4123, 4132, 4213, 4231, 4312, 4321

 13                         };

 14 

 15 

 16 struct Team

 17 {

 18     int id;

 19     int qiu;

 20     int point;

 21 };

 22 

 23 bool vis[30];

 24 int game[5][5], gori[5][5];   // 0=i输j 1=i平j 3=i赢j

 25 Team tm[5], ori[5];

 26 int N, cnt;

 27 

 28 void show()

 29 {

 30     for ( int i = 1; i <= 4; ++i )

 31         printf( "id=%d point=%d qiu=%d\n", i, tm[i].point, tm[i].qiu );

 32     puts("");

 33     return;

 34 }

 35 

 36 void chuli( int i, int j, int c )   //c为队伍i的净赢球量

 37 {

 38     tm[i].point += game[i][j];

 39     tm[j].point += game[j][i];

 40     tm[i].qiu += c;

 41     tm[j].qiu -= c;

 42     return;

 43 }

 44 

 45 void init()

 46 {

 47     memset( vis, false, sizeof(vis) );

 48     memset( game, -1, sizeof( game ) );

 49     memset( tm, 0, sizeof(tm) );

 50     for ( int i = 0; i < N; ++i )

 51     {

 52         int a, b, c, d;

 53         scanf( "%d%d%d%d", &a, &b, &c, &d );

 54         if ( c == d )

 55         {

 56             game[a][b] = game[b][a] = 1;

 57         }

 58         else if ( c > d )

 59         {

 60             game[a][b] = 3;

 61             game[b][a] = 0;

 62         }

 63         else

 64         {

 65             game[a][b] = 0;

 66             game[b][a] = 3;

 67         }

 68         chuli( a, b, c - d );

 69     }

 70     return;

 71 }

 72 

 73 bool DFS( int *tmp, int cur )

 74 {

 75     if ( cur <= 0 )

 76     {

 77         /*

 78         for ( int i = 1; i <= 4; ++i )

 79             printf( "id=%d point=%d qiu=%d\n", tmp[i], tm[ tmp[i] ].point, tm[ tmp[i] ].qiu );

 80         puts("");

 81 */

 82         for ( int i = 1; i < 4; ++i )

 83         {

 84             int u = tmp[i];

 85             int v = tmp[i + 1];

 86             if ( tm[u].point < tm[v].point || ( tm[u].point == tm[v].point && tm[u].qiu < tm[v].qiu ) )

 87                 return false;

 88         }

 89         return true;

 90     }

 91 

 92     int j = tmp[cur];

 93     bool ok = true;

 94     for ( int i = 1; i <= 4; ++i )

 95     {

 96         if ( i == j ) continue;

 97         if ( game[j][i] == -1 )

 98         {

 99             ok = false;

100             for ( int k = -10; k <= 10; ++k )   //枚举净赢球数

101             {

102                 if ( k < 0 )

103                 {

104                     game[j][i] = 0;

105                     game[i][j] = 3;

106                 }

107                 else if ( k == 0 )

108                 {

109                     game[j][i] = 1;

110                     game[i][j] = 1;

111                 }

112                 else

113                 {

114                     game[j][i] = 3;

115                     game[i][j] = 0;

116                 }

117                 chuli( j, i, k );

118 

119                 //puts("*************");

120                 //show();

121 

122                 if ( DFS( tmp, cur ) ) return true;

123 

124                 tm[j].point -= game[j][i];

125                 tm[i].point -= game[i][j];

126                 game[j][i] = -1;

127                 game[i][j] = -1;

128                 tm[i].qiu += k;

129                 tm[j].qiu -= k;

130 

131                 //show();

132                 //puts("==============");

133 

134             }

135         }

136     }

137 

138     if ( ok )

139     {

140         if ( cur != 4 )

141         {

142             int u = tmp[cur];

143             int v = tmp[cur + 1];

144             //printf("%d %d %d %d\n",tm[u].point, tm[v].point, tm[u].qiu, tm[v].qiu );

145             if ( tm[u].point < tm[v].point || ( tm[u].point == tm[v].point && tm[u].qiu < tm[v].qiu ) )

146                 return false;

147             else

148             {

149                 if ( DFS( tmp, cur - 1 ) ) return true;

150             }

151         }

152         else

153         {

154             if ( DFS( tmp, cur - 1 ) ) return true;

155         }

156     }

157     return false;

158 }

159 

160 void solved()

161 {

162     for ( int i = 1; i <= 4; ++i )

163     {

164         ori[i] = tm[i];

165         for ( int j = 1; j <= 4; ++j )

166             gori[i][j] = game[i][j];

167     }

168 

169     int temp[5];

170     for ( int i = 0; i < 24; ++i )

171     {

172         int nn = permuta[i];

173         for ( int j = 4; j > 0; --j )

174         {

175             temp[j] = nn % 10;

176             nn /= 10;

177         }

178 /*

179         for ( int j = 1; j <= 4; ++j )

180             printf( "%d ", temp[j] );

181         puts("");

182 */

183         for ( int j = 1; j <= 4; ++j )

184         {

185             tm[j] = ori[j];

186             for ( int k = 1; k <= 4; ++k )

187                 game[j][k] = gori[j][k];

188         }

189 

190         if ( DFS( temp, 4 ) )

191         {

192             ++cnt;

193             vis[i] = true;

194         }

195 

196   //      puts("-------");

197     }

198 }

199 

200 int main()

201 {

202     //freopen("s.out", "w", stdout );

203     while ( ~scanf( "%d", &N ) )

204     {

205         init();

206 

207         cnt = 0;

208         solved();

209 

210         int ans[4];

211         printf( "%d\n", cnt );

212         for ( int i = 0; i < 24; ++i )

213             if ( vis[i] )

214             {

215                 int nn = permuta[i];

216                 for ( int j = 3; j >= 0; --j )

217                 {

218                     ans[j] = nn % 10;

219                     nn /= 10;

220                 }

221                 for ( int j = 0; j < 4; ++j )

222                 {

223                     if ( j ) putchar(' ');

224                     printf( "%d", ans[j] );

225                 }

226                 puts("");

227             }

228     }

229     return 0;

230 }