南阳37

  1 /*
  2 将字符串逆序，找出原字符串与该字符串的公共子序列，则其余部分为需要添加的字符
  3 状态转移方程参考最长公共子字符串
  4 http://www.cnblogs.com/qq188380780/p/6678471.html
  5 */
  6 #include<cstdio>
  7 #include<cstring>
  8 #define Max(a,b) ((a)>(b)?(a):(b))
  9 char a[1005],b[1005];
 10 int d[1005][1005];
 11
 12 void solven(int len)
 13 {
 14     int i,j;
 15     for(i=1; i<=len; ++i)
 16     {
 17         for(j=1; j<=len; ++j)
 18         {
 19             if(a[i-1] == b[j-1])
 20                 d[i][j] = d[i-1][j-1] + 1;
 21             else
 22                 d[i][j] = Max(d[i-1][j],d[i][j-1]);
 23         }
 24     }
 25 }
 26
 27 int main()
 28 {
 29     int t,i,j,len;
 30     scanf("%d",&t);
 31     while(t--)
 32     {
 33         memset(d,0,sizeof d);
 34         scanf("%s",a);
 35         len = strlen(a);
 36         for(i=0; i<len; ++i)
 37             b[i] = a[len-1-i];
 38         solven(len);
 39         printf("%d\n",len-d[len][len]);
 40     }
 41     return 0;
 42 }
 43
 44 //空间优化，与最长公共子序列一样
 45 #include<cstdio>
 46 #include<cstring>
 47
 48 char a[1005],b[1005];
 49 int d[1005];
 50
 51 void solven(int len)
 52 {
 53     int i,j,t,temp;
 54     for(i=0; i<len; ++i)
 55     {
 56         for(j=t=0; j<len; ++j)
 57         {
 58             temp = d[j];
 59             if(a[i] == b[j])
 60                 d[j] = t + 1;
 61             else if(d[j] < d[j-1])
 62                 d[j] = d[j-1];
 63             t = temp;
 64         }
 65     }
 66 }
 67
 68 int main()
 69 {
 70     int t,i,j,len;
 71     scanf("%d",&t);
 72     while(t--)
 73     {
 74         memset(d,0,sizeof d);
 75         scanf("%s",a);
 76         len = strlen(a);
 77         for(i=0; i<len; ++i)
 78             b[i] = a[len-1-i];
 79         solven(len);
 80         printf("%d\n",len-d[len-1]);
 81     }
 82     return 0;
 83 }
 84
 85 //另可不必再开数组存逆序，只需要记录位置即可
 86 #include<cstdio>
 87 #include<cstring>
 88
 89 char a[1005];
 90 int d[1005];
 91
 92 void solven(int len)
 93 {
 94     int i,j,t,temp;
 95     for(i=0; i<len; ++i)
 96     {
 97         for(j=t=0; j<len; ++j)
 98         {
 99             temp = d[j];
100             if(a[i] == a[len-1-j])
101                 d[j] = t + 1;
102             else if(d[j] < d[j-1])
103                 d[j] = d[j-1];
104             t = temp;
105         }
106     }
107 }
108
109 int main()
110 {
111     int t,i,j,len;
112     scanf("%d",&t);
113     while(t--)
114     {
115         memset(d,0,sizeof d);
116         scanf("%s",a);
117         len = strlen(a);
118         for(i=0; i<len; ++i)
119             b[i] = a[len-1-i];
120         solven(len);
121         printf("%d\n",len-d[len-1]);
122     }
123     return 0;
124 }