HDU-4681 String 枚举+DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4681

　　题意：给A,B,C三个串，求一个最长的串D，满足D是A和B的subsequence，C是D的substring。。

　　比赛那天把substing搞成了subsequence,,,sd。。。

　　挺水的一题，直接枚举C在A和B串中的位置，当然是最短的位置，然后求两遍A和B的最长公共子序列，一个从前往后，另一个从后往前，然后遍历枚举就可以了，O(n^2)..

 //STATUS:C++_AC_343MS_8164KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const LL MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e50;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 char A[N],B[N],C[N];
 int da[N][],db[N][],f1[N][N],f2[N][N];
 int T;

 void getd(char a[],char c[],int d[][],int &cnt)
 {
     cnt=;
     int i,j,k,len=strlen(a),lenc=strlen(c);
     for(i=;i<len;i++){
         if(a[i]!=c[])continue;
         for(j=i,k=;j<len;j++){
             if(a[j]==c[k]){
                 k++;
                 if(k==lenc)break;
             }
         }
         if(k==lenc){
             d[cnt][]=i;
             d[cnt++][]=j;
         }
     }
 }

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,lena,lenb,lenc,ca=;
     int cnta,cntb,ans;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%s%s%s",A,B,C);
         lena=strlen(A);lenb=strlen(B);lenc=strlen(C);
         getd(A,C,da,cnta);
         getd(B,C,db,cntb);

         for(i=;i<=lena || i<=lenb;i++)
                 f1[i][]=f1[][i]=f2[i][lenb]=f2[lena][i]=;
         for(i=;i<=lena;i++){
             for(j=;j<=lenb;j++){
                 f1[i][j]=Max(f1[i][j-],f1[i-][j]);
                 if(A[i-]==B[j-])f1[i][j]=Max(f1[i][j],f1[i-][j-]+);
             }
         }
         for(i=lena-;i>=;i--){
             for(j=lenb-;j>=;j--){
                 f2[i][j]=Max(f2[i+][j],f2[i][j+]);
                 if(A[i]==B[j])f2[i][j]=Max(f2[i][j],f2[i+][j+]+);
             }
         }
         ans=;
         for(i=;i<cnta;i++){
             for(j=;j<cntb;j++){
                 ans=Max(ans,f1[da[i][]][db[j][]]+f2[da[i][]+][db[j][]+]);
             }
         }
         if(!cnta || !cntb)ans=-lenc;

         printf("Case #%d: %d\n",ca++,ans+lenc);
     }
     return ;
 }