最长公共子序列变形

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

3

USA

USSR

LAILI

MAJNU

SHAHJAHAN

MOMTAJ

Sample Output

Case 1: 5 3

Case 2: 9 40

Case 3: 13 15

题意 ： 两个小问，第一问是求一个最短的长度构成的序列同时包含这两个序列，第二问是求构成这个最短序列的方案数

思路分析 ： 第一问就是用两个串的长度减去 lcs

　　　　　　第二问定义 dp[i][j][k] 表示第一个串用了 i 个字符， 第二个串用了j 个字符，并且当前匹配用去了 k 个字符的方案数

　　　　转移过程类似求 lcs 的过程

代码示例：

#define ll long long

char a[100], b[100];
ll dp[100][100][100];
ll dp2[100][100];
ll lena, lenb, num;

void solve(){

    lena = strlen(a+1);
    lenb = strlen(b+1);
    memset(dp2, 0, sizeof(dp2));
    for(ll i = 1; i <= lena; i++){
        for(ll j = 1; j <= lenb; j++){
            if (a[i] == b[j]) dp2[i][j] = dp2[i-1][j-1]+1;
            else dp2[i][j] = max(dp2[i-1][j], dp2[i][j-1]);
        }
    }
    memset(dp, 0, sizeof(dp));
    dp[0][0][1] = 1;
    num = dp2[lena][lenb];
    for(ll i = 1; i <= lena; i++) dp[i][0][1] = 1;
    for(ll i = 1; i <= lenb; i++) dp[0][i][1] = 1;

    for(ll i = 1; i <= lena; i++){
        for(ll j = 1; j <= lenb; j++){
            for(ll k = 1; k <= num+1; k++){
                if (a[i] == b[j]) dp[i][j][k] = dp[i-1][j-1][k-1];
                else dp[i][j][k] = dp[i-1][j][k]+dp[i][j-1][k];
            }
        }
    }

}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ll t;
    ll cas = 1;

    cin >> t;
    while(t--){
        scanf("%s%s", a+1, b+1);
        solve();
        printf("Case %lld: %lld %lld\n", cas++, lena+lenb-num, dp[lena][lenb][num+1]);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/ccut-ry/p/10296013.html

时间： 2024-11-01 18:20:39

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