USACO 4.3 Letter Game(DFS)

poj1171 是这个题的简化版，我的思路是全排列字母，用tire树把字典里的东西都存起来，写的非常麻烦。。。全排列简单dfs就可以，得到结果 判断此字符串是不是在字典里，分割是不是在字典里。。。写的很繁琐，一直MLE，我看不懂题意，还一直以为是读入问题呢，把最后结果map水了一下。。。终于过了，太恶心了。

  1 /*

  2 ID:cuizhe

  3 LANG: C++

  4 TASK: lgame

  5 */

  6 #include <cstdio>

  7 #include <cmath>

  8 #include <cstring>

  9 #include <cstdlib>

 10 #include <iostream>

 11 #include <string>

 12 #include <map>

 13 #include <queue>

 14 #include <algorithm>

 15 using namespace std;

 16 map<string,int> mp;

 17 int n,t = 0;

 18 int aim[26] = {2,5,4,4,1,6,5,5,1,7,6,3,5,2,3,5,7,2,1,2,4,6,6,7,5,7};

 19 char str[8];

 20 int o[8];

 21 int maxz = 0,inf;

 22 int tire[100000][27];

 23 int flag[100000];

 24 int top;

 25 char pos[100][10];

 26 void build(char *p)

 27 {

 28     int i,root = 0,len;

 29     len = strlen(p);

 30     for(i = 0; i < len; i ++)

 31     {

 32         if(tire[root][p[i]-'a'])

 33             ;

 34         else

 35             tire[root][p[i]-'a'] = t ++;

 36         root = tire[root][p[i]-'a'];

 37     }

 38     flag[root] = 1;

 39 }

 40 int find(char *p)

 41 {

 42     int i,root = 0,len;

 43     len = strlen(p);

 44     for(i = 0; i < len; i ++)

 45     {

 46         if(tire[root][p[i]-'a'])

 47             root = tire[root][p[i]-'a'];

 48         else

 49             break;

 50     }

 51     if(i == len&&flag[root]) return 1;

 52     else return 0;

 53 }

 54 void dfs(int x)

 55 {

 56     int i,j,sum = 0,len;

 57     char ch[8],l[6],r[6];

 58     int w[201];

 59     memset(w,0,sizeof(w));

 60     if(x > n)

 61     {

 62         sum = inf;

 63         len = n;

 64         for(i = 1; i <= len; i ++)

 65         {

 66             for(j = 0; j < len; j ++)

 67             {

 68                 if(o[j] == i)

 69                 {

 70                     ch[i-1] = str[j];

 71                     break;

 72                 }

 73             }

 74         }

 75         ch[len] = '\0';

 76         for(   ; len >= 3; sum -= aim[ch[len-1]-'a'],len--)

 77         {

 78             ch[len] = '\0';

 79             if(find(ch) == 1&&sum >= maxz)

 80             {

 81                 if(sum > maxz)

 82                 {

 83                     maxz = sum;

 84                     top = 0;

 85                     strcpy(pos[top],ch);

 86                     top ++;

 87                 }

 88                 else if(maxz == sum)

 89                 {

 90                     strcpy(pos[top],ch);

 91                     top ++;

 92                 }

 93             }

 94             if(len < 6) continue;

 95             for(i = 0; i < 3; i ++)

 96                 l[i] = ch[i];

 97             l[i] = '\0';

 98             for(; i < len; i ++)

 99                 r[i-3] = ch[i];

100             r[i-3] = '\0';

101             if(strcmp(l,r) < 0)

102             {

103                 if(find(l)&&find(r)&&sum >= maxz)

104                 {

105                     if(sum > maxz)

106                         top = 0;

107                     maxz = sum;

108                     for(i = 0; i < 3; i ++)

109                         pos[top][i] = l[i];

110                     pos[top][3] = ' ';

111                     for(i = i+1,j = 0; r[j]; i ++,j ++)

112                         pos[top][i] = r[j];

113                     pos[top][i] = '\0';

114                     top ++;

115                 }

116             }

117             for(i = 0; i < 4; i ++)

118                 l[i] = ch[i];

119             l[i] = '\0';

120             for(; i < len; i ++)

121                 r[i-4] = ch[i];

122             r[i-4] = '\0';

123             if(strcmp(l,r) < 0)

124             {

125                 if(find(l)&&find(r)&&sum >= maxz)

126                 {

127                     if(sum > maxz)

128                     top = 0;

129                     maxz = sum;

130                     for(i = 0; i < 4; i ++)

131                         pos[top][i] = l[i];

132                     pos[top][4] = ' ';

133                     for(i = i+1,j = 0; r[j]; i ++,j ++)

134                         pos[top][i] = r[j];

135                     pos[top][i] = '\0';

136                     top ++;

137                 }

138             }

139         }

140     }

141     for(i = 0; i < n; i ++)

142     {

143         if(!o[i]&&!w[str[i]])

144         {

145             w[str[i]] = 1;

146             o[i] = x;

147             dfs(x+1);

148             o[i] = 0;

149         }

150     }

151 }

152 int main()

153 {

154     int i,j;

155     char temp;

156     char ch[10];

157     freopen("lgame.dict","r",stdin);

158     while(scanf("%s",ch)!=EOF)

159     {

160         if(strcmp(ch,".") == 0) break;

161         build(ch);

162     }

163     freopen("lgame.in","r",stdin);

164     freopen("lgame.out","w",stdout);

165     scanf("%s",str);

166     n = strlen(str);

167     inf = 0;

168     for(i = 1; i <= n-1; i ++)

169     {

170         for(j = 0; j <= n-i-1; j ++)

171         {

172             if(str[j] > str[j+1])

173             {

174                 temp = str[j];

175                 str[j] = str[j+1];

176                 str[j+1] = temp;

177             }

178         }

179     }

180     for(i = 0; i < n; i ++)

181         inf += aim[str[i]-'a'];

182     dfs(1);

183     printf("%d\n",maxz);

184     for(i = 0; i < top; i ++)

185     {

186         if(!mp[pos[i]])

187         {

188             mp[pos[i]] = 1;

189             printf("%s\n",pos[i]);

190         }

191     }

192     return 0;

193 }