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 }
