【解题报告】uva10405_Longest Common Subsequence（最长公共子序列, dp）

Problem C: Longest Common Subsequence

Sequence 1:                

Sequence 2:                


Given two sequences of characters, print the length of the longest common subsequence of both sequences. For example, the longest common subsequence of the following two sequences:

abcdghaedfhr

is adh of length 3.

Input consists of pairs of lines. The first line of a pair contains the first string and the second line contains the second string. Each string is on a separate line and consists of at most 1,000 characters

For each subsequent pair of input lines, output a line containing one integer number which satisfies the criteria stated above.

Sample input

a1b2c3d4ezz1yy2xx3ww4vvabcdghaedfhrabcdefghijklmnopqrstuvwxyza0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0abcdefghijklmnzyxwvutsrqpoopqrstuvwxyzabcdefghijklmn

Output for the sample input

432614

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Problem Setter: Piotr Rudnicki

题目大意：

输入两个长度不相等且均不超过1000的字符串，求其最长公共子序列的长度。

解题思路：

基础动态规划，设输入的字符串分别为A、B。定义状态dp(i)(j)，表示A串前i个字符与B串前j个字符的最长公共子序列的长度。

状态转移方程：dp(i)(j) = max{ dp(i-1)(j), dp(i)(j-1) | A[i]!=B[j] }

      dp(i)(j) = max{ dp(i-1)(j), dp(i)(j-1), dp(i-1)(j-1)+1 | A[i]=B[j] }

注意输入的字符串中包含空格。

#include <cstdio>#include <cstring>#define max(a,b) (a>b?a:b)char A[1010];char B[1010];int dp[1010][1010];int main(){    //freopen("in.txt","r",stdin);    while(gets(A+1) && gets(B+1)){////gets()的返回值是读入的字符串,错误返回NULL(0)        memset(dp,0,sizeof(dp));        int lenA = strlen(A+1);        int lenB = strlen(B+1);        for(int i=1;i<=lenA;++i){            for(int j=1;j<=lenB;++j){                dp[i][j]=max(dp[i-1][j],dp[i][j-1]);                if(A[i]==B[j]) dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1);            }        }        printf("%d\n",dp[lenA][lenB]);    }    return 0;}