String

题目连接：

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

Description

Given 3 strings A, B, C, find the longest string D which satisfy the following rules:

a) D is the subsequence of A

b) D is the subsequence of B

c) C is the substring of D
Substring here means a consecutive subsequnce.
You need to output the length of D.

Input

The first line of the input contains an integer T(T = 20) which means the number of test cases.
For each test case, the first line only contains string A, the second line only contains string B, and the third only contains string C.
The length of each string will not exceed 1000, and string C should always be the subsequence of string A and string B.
All the letters in each string are in lowercase.

Output

For each test case, output Case #a: b. Here a means the number of case, and b means the length of D.

Sample Input

2
aaaaa
aaaa
aa
abcdef
acebdf
cf

Sample Output

Case #1: 4
Case #2: 3

Hint

题意

给你a,b,c串，你需要找到一个最长的d串

这个d串是ab串的子序列，然后c串是d串的子串

问你这个d串的长度是多少

题解：

数据范围1000，我们显然可以暴力枚举在a串的起始位置和在b串的起始位置

答案就是ansL + lenc + ansR

ansL就是ab串之前的最长公共子序列，ansR就是在匹配之后后面的最长公共子序列

lenc这个就是贪心匹配的最短位置在哪儿

这些东西都可以n^2的复杂度直接预处理

然后暴力去莽一波就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1050;
int dp1[maxn][maxn],dp2[maxn][maxn],l1[maxn],l2[maxn],len1,len2,len3;
char s1[maxn],s2[maxn],s3[maxn];

inline void update( int & x , int v ) { x = max( x , v ) ; }

void pre()
{
    for(int i=1;i<=len1;i++)
    {
        int l=1;
        for(int j=i;j<=len1;j++)
        {
            if(s1[j]==s3[l])l++;
            if(l>len3)
            {
                l1[i]=j;
                break;
            }
        }
    }
    for(int i=1;i<=len2;i++)
    {
        int l=1;
        for(int j=i;j<=len2;j++)
        {
            if(s2[j]==s3[l])l++;
            if(l>len3)
            {
                l2[i]=j;
                break;
            }
        }
    }
    for(int i=0 ;i< len1;i++)
    {
        for(int j=0 ;j< len2;j++)
        {
            if(s1[i + 1] == s2[j + 1] ) update( dp1[i+1][j+ 1] , dp1[i][j] + 1 );
            update(dp1[i + 1][j] , dp1[i][j] );
            update(dp1[i][j + 1] , dp1[i][j] );
        }
    }
    reverse(s1+1,s1+1+len1);
    reverse(s2+1,s2+1+len2);

    for(int i=0 ;i< len1;i++)
    {
        for(int j=0 ;j< len2;j++)
        {
            if(s1[i + 1] == s2[j + 1] ) update( dp2[i+1][j+ 1] , dp2[i][j] + 1 );
            update(dp2[i + 1][j] , dp2[i][j] );
            update(dp2[i][j + 1] , dp2[i][j] );
        }
    }
}
void solve(int cas)
{
    int ans = 0;
    for(int i=1;i<=len1;i++)
    {
        for(int j=1;j<=len2;j++)
        {
            if(l1[i]==-1||l2[j]==-1)continue;
            ans = max(dp1[i-1][j-1]+dp2[len1-l1[i]][len2-l2[j]]+len3,ans);
        }
    }
    printf("Case #%d: %d\n",cas,ans);
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        memset(dp1,0,sizeof(dp1));
        memset(dp2,0,sizeof(dp2));
        memset(l1,-1,sizeof(l1));
        memset(l2,-1,sizeof(l2));
        scanf("%s",s1+1);
        scanf("%s",s2+1);
        scanf("%s",s3+1);
        len1=strlen(s1+1),len2=strlen(s2+1),len3=strlen(s3+1);
        pre();
        solve(i);
    }
}