题目地址:
https://www.acwing.com/problem/content/899/
给定两个长度分别为 N N N和 M M M的字符串 A A A和 B B B,求既是 A A A的子序列又是 B B B的子序列的字符串长度最长是多少。
数据范围:
1 ≤ N , M ≤ 1000 1\le N,M\le 1000 1≤N,M≤1000
思路是动态规划。设 f [ i ] [ j ] f[i][j] f[i][j]是 A A A的前 i i i个字符和 B B B的前 j j j个字符的最长公共子序列的长度。当 A [ i ] ≠ B [ j ] A[i]\ne B[j] A[i]=B[j]的时候,显然 f [ i ] [ j ] = max { f [ i − 1 ] [ j ] , f [ i ] [ j − 1 ] } f[i][j]=\max\{f[i-1][j],f[i][j-1]\} f[i][j]=max{ f[i−1][j],f[i][j−1]};当 A [ i ] = B [ j ] A[i]=B[j] A[i]=B[j]的时候 f [ i ] [ j ] = 1 + f [ i − 1 ] [ j − 1 ] f[i][j]=1+f[i-1][j-1] f[i][j]=1+f[i−1][j−1],原因是,首先显然 f [ i ] [ j ] ≥ 1 + f [ i − 1 ] [ j − 1 ] f[i][j]\ge 1+f[i-1][j-1] f[i][j]≥1+f[i−1][j−1],其次,如果 f [ i ] [ j ] > 1 + f [ i − 1 ] [ j − 1 ] f[i][j]> 1+f[i-1][j-1] f[i][j]>1+f[i−1][j−1]的话,如果取到 f [ i ] [ j ] f[i][j] f[i][j]最后一个相等的字符是 A [ i ] A[i] A[i]和 B [ j ] B[j] B[j],那么 f [ i ] [ j ] = 1 + f [ i − 1 ] [ j − 1 ] f[i][j]=1+f[i-1][j-1] f[i][j]=1+f[i−1][j−1],矛盾;如果不是,则有 f [ i ] [ j ] = max { f [ i − 1 ] [ j ] , f [ i ] [ j − 1 ] } f[i][j]=\max\{f[i-1][j],f[i][j-1]\} f[i][j]=max{ f[i−1][j],f[i][j−1]},如果是 f [ i − 1 ] [ j ] > 1 + f [ i − 1 ] [ j − 1 ] f[i-1][j]>1+f[i-1][j-1] f[i−1][j]>1+f[i−1][j−1]的话,同样考虑取到 f [ i − 1 ] [ j ] f[i-1][j] f[i−1][j]的最后一个字符,如果用到了 t [ j ] t[j] t[j]就会得出 1 + f [ i − 1 ] [ j − 1 ] > 1 + f [ i − 1 ] [ j − 1 ] 1+f[i-1][j-1]>1+f[i-1][j-1] 1+f[i−1][j−1]>1+f[i−1][j−1]矛盾;如果没用到,就会得出 f [ i − 1 ] [ j − 1 ] > 1 + f [ i − 1 ] [ j − 1 ] f[i-1][j-1]>1+f[i-1][j-1] f[i−1][j−1]>1+f[i−1][j−1]也矛盾。所以有 f [ i ] [ j ] = 1 + f [ i − 1 ] [ j − 1 ] f[i][j]=1+f[i-1][j-1] f[i][j]=1+f[i−1][j−1]。代码如下:
#include <iostream>
using namespace std;
const int N = 1010;
string s, t;
int n, m;
int f[N][N];
int main() {
cin >> n >> m;
cin >> s >> t;
s = ' ' + s;
t = ' ' + t;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
if (s[i] == t[j]) f[i][j] = 1 + f[i - 1][j - 1];
else f[i][j] = max(f[i - 1][j], f[i][j - 1]);
cout << f[n][m] << endl;
return 0;
}
时空复杂度 O ( N M ) O(NM) O(NM)。