（最长公共子序列+推导）Love Calculator （lightOJ 1013）

http://www.lightoj.com/volume_showproblem.php?problem=1013

 

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：

给你两个串， 让你求出这两个串能组成的最短串的长度，以及最短串多少种不同的串（第三个串中保证有最长公共字串）

 

最长公共子序列， 在求最长公共子序列的过程中加上每个串有多少种组成方法

 

dp[i][j] 代表第一个串的前i个和第二个串的前j个的最长公共子序列

num[i][j]代表第一个串的前i个和第二个串的钱j个能组成多少种不同的串

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>

using namespace std;

#define N 1100

#define met(a,b) (memset(a,b,sizeof(a)))
typedef long long LL;

int a[20][20], dp[50][50];
LL num[50][50];

int main()
{
    int T, iCase=1;

    scanf("%d", &T);

    while(T--)
    {
        char s1[50], s2[50];
        int i, j, len1, len2;

        met(dp, 0);
        met(num, 0);
        scanf("%s%s", s1, s2);
        len1 = strlen(s1);
        len2 = strlen(s2);


        for(i=0; i<=len1; i++)
            num[i][0] = 1;
        for(i=0; i<=len2; i++)
            num[0][i] = 1;

        for(i=1; i<=len1; i++)
        for(j=1; j<=len2; j++)
        {
            if(s1[i-1]==s2[j-1])
            {
                dp[i][j] = dp[i-1][j-1] + 1;
                num[i][j] += num[i-1][j-1];
            }
            else
            {
                if(dp[i-1][j]>dp[i][j-1])
                {
                    dp[i][j] = dp[i-1][j];
                    num[i][j] = num[i-1][j];
                }
                else if(dp[i-1][j]<dp[i][j-1])
                {
                    dp[i][j] = dp[i][j-1];
                    num[i][j] = num[i][j-1];
                }
                else
                {
                    dp[i][j] = dp[i-1][j];
                    num[i][j] = num[i-1][j] + num[i][j-1];
                }
            }
        }

        printf("Case %d: %d %lld\n", iCase++, len1+len2-dp[len1][len2], num[len1][len2]);
    }
    return 0;
}

/**
3
USA
USSR
LAILI
MAJNU
SHAHJAHAN
MOMTAJ
*/

 

自己写完后， 搜题解看到的另一种写法， 由于dp刚入门， 就学习一下思想

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。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>

using namespace std;

#define N 1100

#define met(a,b) (memset(a,b,sizeof(a)))
typedef long long LL;

LL dp[100][50][50];

int main()
{
    int T, iCase=1;

    scanf("%d", &T);

    while(T--)
    {
        char s1[50], s2[50];
        int i, j, k, len1, len2;

        met(dp, 0);
        scanf("%s%s", s1, s2);
        len1 = strlen(s1);
        len2 = strlen(s2);


        for(i=0; i<=len1; i++)
            dp[i][i][0] = 1;
        for(i=0; i<=len2; i++)
            dp[i][0][i] = 1;

        for(i=1; i<=len1+len2; i++)
        {
            for(j=1; j<=len1; j++)
            for(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(k=1; k<=len1+len2; k++)
            if(dp[k][len1][len2]) break;

        printf("Case %d: %d %lld\n", iCase++, k, dp[k][len1][len2]);
    }
    return 0;
}

/**
3
USA
USSR
LAILI
MAJNU
SHAHJAHAN
MOMTAJ
*/