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=1010;
const int INF=0x3f3f3f3f;
const LL MOD=1000000007,STA=8000010;
const LL LNF=1LL<<55;
const double EPS=1e-9;
const double OO=1e50;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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][2],db[N][2],f1[N][N],f2[N][N];
int T;

void getd(char a[],char c[],int d[][2],int &cnt)
{
    cnt=0;
    int i,j,k,len=strlen(a),lenc=strlen(c);
    for(i=0;i<len;i++){
        if(a[i]!=c[0])continue;
        for(j=i,k=0;j<len;j++){
            if(a[j]==c[k]){
                k++;
                if(k==lenc)break;
            }
        }
        if(k==lenc){
            d[cnt][0]=i;
            d[cnt++][1]=j;
        }
    }
}

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j,lena,lenb,lenc,ca=1;
    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=0;i<=lena || i<=lenb;i++)
                f1[i][0]=f1[0][i]=f2[i][lenb]=f2[lena][i]=0;
        for(i=1;i<=lena;i++){
            for(j=1;j<=lenb;j++){
                f1[i][j]=Max(f1[i][j-1],f1[i-1][j]);
                if(A[i-1]==B[j-1])f1[i][j]=Max(f1[i][j],f1[i-1][j-1]+1);
            }
        }
        for(i=lena-1;i>=0;i--){
            for(j=lenb-1;j>=0;j--){
                f2[i][j]=Max(f2[i+1][j],f2[i][j+1]);
                if(A[i]==B[j])f2[i][j]=Max(f2[i][j],f2[i+1][j+1]+1);
            }
        }
        ans=0;
        for(i=0;i<cnta;i++){
            for(j=0;j<cntb;j++){
                ans=Max(ans,f1[da[i][0]][db[j][0]]+f2[da[i][1]+1][db[j][1]+1]);
            }
        }
        if(!cnta || !cntb)ans=-lenc;

        printf("Case #%d: %d
",ca++,ans+lenc);
    }
    return 0;
}