light oj 1013 - Love Calculator（DP求两个字符串的构造序列方法）

1013 - Love Calculator

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Yes, you are developing a 'Love calculator'. The software would be quite complex such that nobody could crack the exact behavior of the software.

So, given two names your software will generate the percentage of their 'love' according to their names. The software requires the following things:

1.                  The length of the shortest string that contains the names as subsequence.

2.                   Total number of unique shortest strings which contain the names as subsequence.

Now your task is to find these parts.

Input

Input starts with an integer T (≤ 125), denoting the number of test cases.

Each of the test cases consists of two lines each containing a name. The names will contain no more than 30 capital letters.

Output

For each of the test cases, you need to print one line of output. The output for each test case starts with the test case number, followed by the shortest length of the string and the number of unique strings that satisfies the given conditions.

You can assume that the number of unique strings will always be less than 263. Look at the sample output for the exact format.

Sample Input	Output for Sample Input
3 USA USSR LAILI MAJNU SHAHJAHAN MOMTAJ	Case 1: 5 3 Case 2: 9 40 Case 3: 13 15

题目大意：

 　　给你两个字符串A,B 要求一个最短的字符串C，使得A,B同时为C的子串。 问C最短长度是多少？ C有多少种？

题目分析：

　　做这道题目的时候自己并没有推出来，看了网上的题解。

1.dp[C串的长度][包含A的字符个数][包含B的字符个数] = 种类数

状态转移：如果 A[i] == B[j] 那么 dp[k][i][j] = dp[k-1][i-1][j-1]. 就是说我最后一个字符是相同的那么我只要放一个就可以了。

　　　　　如果 A[i] !=  B[j] 那么 dp[k][i][j] = dp[k-1][i-1][j] + dp[k-1][i][j-1].最后一个字符我们要么放A[i] 要么放 B[j] 就这两种情况了。

然后关于找最短的，就可以在 dp[k][lenA][lenB] 种找到最小的k即可

用DP 来处理方法的问题很经典，但是开了三维数组，不好想，从全局考虑的话可能容易理解一点

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

const int N = 100;

char s1
, s2
;

long long  dp[2*N]

;

int main()

{

    int t,  ncase=1;

    scanf("%d", &t);

    while(t--)

    {

        scanf("%s %s",s1, s2);

        memset(dp,0,sizeof(dp));

        int len1=strlen(s1);

        int len2=strlen(s2);

        for(int i=0;i<=len1;i++) dp[i][i][0]=1;

        for(int i=0;i<=len2;i++) dp[i][0][i]=1;

        for(int i=1;i<=len1+len2;i++)

        {

            for(int j=1;j<=len1;j++)

            {

                for(int k=1;k<=len2;k++)

                {

                    if(s1[j-1]==s2[k-1])

                        dp[i][j][k]=dp[i-1][j-1][k-1];

                    else

                        dp[i][j][k]=dp[i-1][j-1][k]+dp[i-1][j][k-1];

                }

            }

        }

        for(int i=1;i<=len1+len2;i++)

        {

            if(dp[i][len1][len2])

            {

                printf("Case %d: %d %lld\n",ncase++,i,dp[i][len1][len2]);

                break;

            }

        }

    }

    return 0;

}