标签：eve could def lov long form 过程 ati cal
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:
Now your task is to find these parts.
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.
OutputFor 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 Input3
USA
USSR
LAILI
MAJNU
SHAHJAHAN
MOMTAJ
Sample OutputCase 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;
}
标签：eve could def lov long form 过程 ati cal
原文地址：https://www.cnblogs.com/ccut-ry/p/10296013.html
