[九度OJ]1042Coincidence

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：2852

解决：1533

题目描述：

Find a longest common subsequence of two strings.

输入：

First and second line of each input case contain two strings of lowercase character a…z. There are no spaces before, inside or after the strings. Lengths of strings do not exceed 100.

输出：

For each case, output k – the length of a longest common subsequence in one line.

样例输入：

abcdcxbydz

样例输出：

2

来源：

2008年上海交通大学计算机研究生机试真题

经典的LCS问题

LCS问题具有最优子结构性质，可以采用动态规划解决，递推公式如下：

a[i][j] = 0,                                      i=0 或 j = 0

             a[i-1][j-1]+1,                    i>0 且 j>0 且 s1[i-1]=s2[j-1]

             max(a[i-1][j], a[i][j-1]),         i>0 且 j>0 且 s1[i-1]!=s2[j-1]

#include <cstdio>#include <cstdlib>#include <iostream>using namespace std;int LCS(const string s1, const string s2) {int n = s1.length()+1;int m = s2.length()+1;int **a = new int*[n];for (int i = 0; i < n; ++i) {a[i] = new int[m];}for (int i = 0; i < n; ++i) {a[i][0] = 0;}for (int i = 0; i < m; ++i) {a[0][i] = 0;}for (int i = 0; i < s1.length(); ++i) {for (int j = 0; j < s2.length(); ++j) {if (s1[i] == s2[j]) {a[i+1][j+1] = a[i][j]+1;}else {a[i+1][j+1] = a[i+1][j]>a[i][j+1] ? a[i+1][j]:a[i][j+1];}}}int ans = a[s1.length()][s2.length()];free(a);return ans;}int main(int argc, char const *argv[]) {string s1, s2;while (cin >> s1 >> s2) {cout << LCS(s1, s2) << endl;}return 0;}