UVA 10405 - Longest Common Subsequence

                                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

s 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

题意：求最长公共子序列。

题目大意：求最长公共子序列设d[i][j]为A1,A2...Ai和B1,B2...Bj的LCS长度，则的d[i][j]=max{d[i-1][j],d[i][j-1]}，如果A[i]=B[j]，d[i][j]=max{d[i][j],d[i-1][j-1]+1}，
#include<stdio.h>#include<string.h>int dp[1005][1005];char s1[1005],s2[1005];int max(int x,int y){return x>y?x:y;}int main(){int len1,len2,i,j;while(gets(s1)!=NULL){gets(s2);len1=strlen(s1);len2=strlen(s2);for(i=0;i<=len1;i++)for(j=0;j<=len2;j++)dp[i][j]=0;for( i=1;i<=len1;i++)for(j=1;j<=len2;j++){dp[i][j]=max(dp[i-1][j],dp[i][j-1]);if(s1[i-1]==s2[j-1])dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1);}printf("%d\n",dp[len1][len2]);}return 0;}