HDOJ1159 Common Subsequence(dp)
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给出两个字符串, 问你最长公共子序列有多长.
lcs模板题目, 摘出求解过程:
第一步:先计算最长公共子序列的长度。
第二步:根据长度,然后通过回溯求出最长公共子序列。
设一个C[i,j]: 保存Xi与Yj的LCS的长度。
递推方程为:
AC代码:
#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;char s1[10005], s2[10005];int dp[10005][10005] = {0};int main(int argc, char const *argv[]){while(scanf("%s%s", s1, s2) != EOF) {int len1 = strlen(s1), len2 = strlen(s2);for(int i = 0; i < len1; ++i)for(int j = 0; j < len2; ++j)if(s1[i] == s2[j]) dp[i + 1][j + 1] = dp[i][j] + 1;else dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j]);printf("%d\n", dp[len1][len2]);}return 0;}
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