杭电OJ Common Subsequence 需要二刷 *仍然是公共子串

基本思想：

还是要注意索引问题；

 

关键点：

无；

 

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;

vector<vector<int>> dp;

int main() {
	string a, b;
	while (cin >> a >> b) {
		int lena = a.size()+1;
		int lenb = b.size()+1;
		int len = max(lena, lenb) + 2;
		dp.resize(len);
		for (int i = 0; i < len; i++) {
			dp[i].resize(len);
		}
		for (int i = 0; i < len; i++)
			dp[i][0] = dp[0][i] = 0;
		for (int i = 1; i < lena; i++) {
			for (int j = 1; j < lenb; j++) {
				if (a[i-1] == b[j-1]) {
					dp[i][j] = dp[i - 1][j - 1] + 1;
				}
				else {
					dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
				}
			}
		}
		cout << dp[lena - 1][lenb - 1] << endl;
	}
}