LightOJ1013---Love Calculator (dp)

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

$f[i][j] 表示第一个串匹配到i,第二个串匹配到j需要的最短长度$
$dp[i][j] 表示f[i][j]长度下的方案数$
转移很简单
但是有一个地方要注意:
如果$A[i] == B[j]$,那么方案数计算的时候,只能从$f[i - 1][j - 1]$处转移到, 如果从$f[i - 1][j] f[i][j - 1]$转移,会重复

/*************************************************************************
    > File Name: LightOJ1013.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年06月07日 星期日 21时41分25秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

string A, B;
LL dp[50][50];
int f[50][50];

int main() {
    int t, icase = 1;
    scanf("%d", &t);
    while (t--) {
        cin >> A >> B;
        int n = A.length();
        int m = B.length();
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
        memset(f, inf, sizeof(f));
        f[0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            f[i][0] = i;
            dp[i][0] = 1;
        }
        for (int j = 1; j <= m; ++j) {
            f[0][j] = j;
            dp[0][j] = 1;
        }
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                if (f[i][j] > f[i - 1][j] + 1) {
                    f[i][j] = f[i - 1][j] + 1;
                }
                if (f[i][j] > f[i][j - 1] + 1) {
                    f[i][j] = f[i][j - 1] + 1;
                }
                if (A[i - 1] == B[j - 1]) {
                    if (f[i][j] > f[i - 1][j - 1] + 1) {
                        f[i][j] = f[i - 1][j - 1] + 1;
                    }
                }
                if (f[i][j] == f[i - 1][j] + 1 && A[i - 1] != B[j - 1]) {
                    dp[i][j] += dp[i - 1][j];
                }
                if (f[i][j] == f[i][j - 1] + 1 && A[i - 1] != B[j - 1]) {
                    dp[i][j] += dp[i][j - 1];
                }
                if (A[i - 1] == B[j - 1] && f[i][j] == f[i - 1][j - 1] + 1) {
                    dp[i][j] += dp[i - 1][j - 1];
                }

            }
        }
        printf("Case %d: %d %lld\n", icase++, f[n][m], dp[n][m]);
    }
}