lightoj 1013 - Love Calculator 【LCS 变形】_lightoj

本文链接：https://blog.youkuaiyun.com/chenzhenyu123456/article/details/51088044

本文针对lightoj1013-LoveCalculator题目进行了解析，介绍了如何通过动态规划算法找到同时包含两个字符串的最短字符串及其数量的方法。

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1013 - Love Calculator
PDF (English) Statistics Forum
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:

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

题意：给定两个串a、b，让你构造一个最短的串包含a和b。问你最短的长度以及可以构造出的不同串的数目。

思路： $dp[i][j]$ 为构造 $a[1]-a[i]$ 和 $b[1]-b[j]$ 的最短长度， $ans[i][j]$ 表示在该状态下的方案数。

$(1) a[i] == b[j]$
$dp[i][j] = dp[i-1][j-1]+1$
$ans[i][j] = ans[i-1][j-1]$

$(2) a[i] != b[j]$
$1、dp[i-1][j] < dp[i][j-1]$
$dp[i][j] = dp[i-1][j] + 1\;\;ans[i][j] = ans[i-1][j]$
$2、dp[i-1][j] > dp[i][j-1]$
$dp[i][j] = dp[i][j-1] + 1\;\; ans[i][j] = ans[i][j-1]$
$3、dp[i-1][j] == dp[i][j-1]$
$dp[i][j] = dp[i][j-1] + 1 \;\; ans[i][j] = ans[i-1][j] + ans[i][j-1]$

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
#define fi first
#define se second
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MOD = 1e4 + 7;
const int MAXN = 1e8;
void add(LL &x, LL y) { x += y; x %= MOD; }
int dp[40][40];
LL ans[40][40];
char a[40], b[40];
int main()
{
    int t, kcase = 1; scanf("%d", &t);
    while(t--) {
        scanf("%s%s", a+1, b+1);
        int la = strlen(a+1);
        int lb = strlen(b+1); CLR(ans, 0);
        for(int i = 1; i <= la; i++) {
            dp[i][0] = i; ans[i][0] = 1;
        }
        for(int j = 1; j <= lb; j++) {
            dp[0][j] = j; ans[0][j] = 1;
        }
        dp[0][0] = 0; ans[0][0] = 1;
        for(int i = 1; i <= la; i++) {
            for(int j = 1; j <= lb; j++) {
                if(a[i] == b[j]) {
                    dp[i][j] = dp[i-1][j-1] + 1;
                }
                else {
                    dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + 1;
                }

                if(a[i] == b[j]) {
                    if(dp[i-1][j-1] + 1 == dp[i][j]) {
                        ans[i][j] = ans[i-1][j-1];
                    }
                }
                else {
                    if(dp[i-1][j] + 1 == dp[i][j]) {
                        ans[i][j] += ans[i-1][j];
                    }
                    if(dp[i][j-1] + 1 == dp[i][j]) {
                        ans[i][j] += ans[i][j-1];
                    }
                }
            }
        }
        printf("Case %d: %d %lld\n", kcase++, dp[la][lb], ans[la][lb]);
    }
    return 0;
}