51nod 1092 回文字符串 (dp)

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1092

这个题是poj-3280的简化版，这里只可以增加字符，设 dp[i][j] 为把以i开头j结尾的子串变为回文串的最少次数，

if(s[i]==s[j])  dp[i][j]=dp[i+1][j-1];

else dp[i][j]=min(dp[i+1][j],dp[i][j-1])+1;

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>
 #pragma comment(linker, "/STACK:102400000,102400000")
 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1000000000
 #define N 1010
 #define mod 1000000000
 using namespace std;

 int dp[N][N];
 char s[N];

 int main()
 {
     //Read();
     scanf("%s",s);
     int m=strlen(s);
     memset(dp,,sizeof(dp));
     for(int i=m-;i>=;i--)
     {
         for(int j=i+;j<m;j++)
         {
                 if(s[i]==s[j]) dp[i][j]=dp[i+][j-];
                 else dp[i][j]=min(dp[i+][j],dp[i][j-])+;
                 //printf("%d\n",dp[i][j]);
         }
     }
     printf("%d\n",dp[][m-]);
     return ;
 }

还可以转化为最长公共子序列的问题求解，求最少变成回文串的操作次数，就是求有几个字符与对应的字符不一样，那么只要我把原字符翻转，然后求最长公共子序列，

用字符串长度减去公共子序列长度即可求解。

因为dp[i+1]计算时只用到了dp[i],dp[i+1],所以可以结合奇偶性用滚动数组优化空间。

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>
 #pragma comment(linker, "/STACK:102400000,102400000")
 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1000000000
 #define N 1010
 #define mod 1000000000
 using namespace std;

 int dp[][N];
 char s[N],ss[N];

 int main()
 {
     //Read();
     scanf("%s",s);
     int m=strlen(s);
     for(int i=;i<m;i++)
         ss[i]=s[m-i-];
     ss[m]='\0';
     memset(dp,,sizeof(dp));
     for(int i=;i<m;i++)
     {
         for(int j=;j<m;j++)
         {
             if(s[i]==ss[j]) dp[(i+) & ][j+]=dp[i & ][j]+;
             else dp[(i+) & ][j+]=max(dp[i & ][j+],dp[(i+) & ][j]);
             //printf("%d\n",dp[i+1][j+1]);
         }
     }
    // printf("%d\n",dp[m][m]);
     printf("%d\n",m-dp[m&][m]);
     return ;
 }