HDU-4632 Palindrome subsequence 区间DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4632

　　区间DP，f[i][j]表示[i,j]区间回文字串的个数。f[i][j]=f[i+1][j]+f[i][j-1]-f[i+1][j-1]+s[i]==s[j]?f[i-1]+f[j-1]+1:0 。

 1 //STATUS:C++_AC_281MS_4188KB

 2 #include <functional>

 3 #include <algorithm>

 4 #include <iostream>

 5 //#include <ext/rope>

 6 #include <fstream>

 7 #include <sstream>

 8 #include <iomanip>

 9 #include <numeric>

10 #include <cstring>

11 #include <cassert>

12 #include <cstdio>

13 #include <string>

14 #include <vector>

15 #include <bitset>

16 #include <queue>

17 #include <stack>

18 #include <cmath>

19 #include <ctime>

20 #include <list>

21 #include <set>

22 #include <map>

23 using namespace std;

24 //#pragma comment(linker,"/STACK:102400000,102400000")

25 //using namespace __gnu_cxx;

26 //define

27 #define pii pair<int,int>

28 #define mem(a,b) memset(a,b,sizeof(a))

29 #define lson l,mid,rt<<1

30 #define rson mid+1,r,rt<<1|1

31 #define PI acos(-1.0)

32 //typedef

33 typedef __int64 LL;

34 typedef unsigned __int64 ULL;

35 //const

36 const int N=1010;

37 const int INF=0x3f3f3f3f;

38 const int MOD=10007,STA=8000010;

39 const LL LNF=1LL<<60;

40 const double EPS=1e-8;

41 const double OO=1e15;

42 const int dx[4]={-1,0,1,0};

43 const int dy[4]={0,1,0,-1};

44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

45 //Daily Use ...

46 inline int sign(double x){return (x>EPS)-(x<-EPS);}

47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

50 template<class T> inline T Min(T a,T b){return a<b?a:b;}

51 template<class T> inline T Max(T a,T b){return a>b?a:b;}

52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

56 //End

57 

58 char s[N];

59 int f[N][N];

60 int n;

61 

62 int main(){

63  //   freopen("in.txt","r",stdin);

64     int Ca=1,T,i,j;

65     scanf("%d",&T);

66     while(T--)

67     {

68         scanf("%s",s);

69         n=strlen(s);

70         for(j=1;j<=n;j++){

71             f[j][j]=1;

72             for(i=j-1;i>=1;i--){

73                 f[i][j]=(f[i+1][j]+f[i][j-1]-f[i+1][j-1])%MOD;

74                 if(s[i-1]==s[j-1])f[i][j]=(f[i][j]+f[i+1][j-1]+1)%MOD;

75             }

76         }

77 

78         printf("Case %d: %d\n",Ca++,(f[1][n]+MOD)%MOD);

79     }

80     return 0;

81 }