CF 346B. Lucky Common Subsequence(DP+KMP)

这题确实很棒。。又是无想法。。其实是AC自动机+DP的感觉，但是只有一个串，用kmp就行了。

dp[i][j][k]，k代表前缀为virus[k]的状态，len表示其他所有状态串，处理出Ac[len][26]数组来，DP就可以了。状态转移那里一直没想清楚，wa了很多次，记录路径倒是不复杂，瞎搞搞就行。

  1 #include<iostream>

  2 #include<cstring>

  3 #include<cstdio>

  4 #include<cmath>

  5 #include<algorithm>

  6 using namespace std;

  7 char s1[103],s2[103],virus[103];

  8 int dp[103][103][103];

  9 int pre[103][103][103];

 10 int pre1[103][103][103];

 11 int Ac[111][31];

 12 int next[103];

 13 char ans[111];

 14 void kmp()

 15 {

 16     int i,j,len,temp;

 17     len = strlen(virus);

 18     next[0] = -1;

 19     j = -1;

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

 21     {

 22         while(j >= 0&&virus[j+1] != virus[i])

 23             j = next[j];

 24         if(virus[j+1] == virus[i]) j ++;

 25         next[i] = j;

 26     }

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

 28     {

 29         for(j = 0; j < 26; j ++)

 30         {

 31             temp = i;

 32             while(temp >= 0&&virus[temp+1] != 'A'+j)

 33                 temp = next[temp];

 34             if(virus[temp+1] == 'A' + j) temp ++;

 35             if(temp == -1)

 36                 Ac[i][j] = len;

 37             else

 38                 Ac[i][j] = temp;

 39         }

 40     }

 41     for(i = 0; i < 26; i ++)

 42     {

 43         if(i + 'A' == virus[0])

 44             Ac[len][i] = 0;

 45         else

 46             Ac[len][i] = len;

 47     }

 48 }

 49 int main()

 50 {

 51     int i,j,k,len1,len2,len,maxz,a,b,kk;

 52     scanf("%s%s%s",s1,s2,virus);

 53     len1 = strlen(s1);

 54     len2 = strlen(s2);

 55     len = strlen(virus);

 56     kmp();

 57     for(i = 1; i <= len1; i ++)

 58     {

 59         for(j = 1; j <= len2; j ++)

 60         {

 61             for(k = 0; k <= len; k ++)

 62             {

 63                 if(k == len-1) continue;

 64                 if(dp[i][j][k] < dp[i-1][j][k])

 65                 {

 66                     dp[i][j][k] = dp[i-1][j][k];

 67                     pre[i][j][k] = 2;

 68                     pre1[i][j][k] = k;

 69                 }

 70                 if(dp[i][j][k] < dp[i][j-1][k])

 71                 {

 72                     dp[i][j][k] = dp[i][j-1][k];

 73                     pre[i][j][k] = 3;

 74                     pre1[i][j][k] = k;

 75                 }

 76                 if(s1[i-1] == s2[j-1])

 77                 {

 78                     if(Ac[k][s1[i-1]-'A'] == len-1) continue;

 79                     else if(dp[i][j][Ac[k][s1[i-1]-'A']] < dp[i-1][j-1][k] + 1)

 80                     {

 81                         dp[i][j][Ac[k][s1[i-1]-'A']] = dp[i-1][j-1][k] + 1;

 82                         pre[i][j][Ac[k][s1[i-1]-'A']] = 1;

 83                         pre1[i][j][Ac[k][s1[i-1]-'A']] = k;

 84                     }

 85                 }

 86             }

 87         }

 88     }

 89     maxz = 0;

 90     for(i = 1; i <= len1; i ++)

 91     {

 92         for(j = 1; j <= len2; j ++)

 93         {

 94             for(k = 0; k <= len; k ++)

 95             {

 96                 if(maxz < dp[i][j][k])

 97                 {

 98                     maxz = dp[i][j][k];

 99                     a = i;

100                     b = j;

101                     kk = k;

102                 }

103             }

104         }

105     }

106     if(maxz == 0)

107     {

108         printf("0\n");

109         return 0;

110     }

111     int num = 0;

112     //printf("%d\n",maxz);

113     while(a != 0&&b != 0)

114     {

115         if(pre[a][b][kk] == 1)

116         {

117             ans[num++] = s1[a-1];

118             kk = pre1[a][b][kk];

119             a --;

120             b --;

121         }

122         else if(pre[a][b][kk] == 2)

123         {

124             kk = pre1[a][b][kk];

125             a --;

126         }

127         else if(pre[a][b][kk] == 3)

128         {

129             kk = pre1[a][b][kk];

130             b --;

131         }

132         else

133             break;

134     }

135     for(i = num-1; i >= 0; i --)

136     {

137         printf("%c",ans[i]);

138     }

139     printf("\n");

140     return 0;

141 }