UVA 10405 Longest Common Subsequence （动态规划 LCS）

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:

abcdgh
aedfhr

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

a1b2c3d4e
zz1yy2xx3ww4vv
abcdgh
aedfhr
abcdefghijklmnopqrstuvwxyz
a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0
abcdefghijklmnzyxwvutsrqpo
opqrstuvwxyzabcdefghijklmn

Output for the sample input

4
3
26
14

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}，边界条件是最外层为0
时间复杂度为O(nm),n、m分别为序列A、B的长度

# include<stdio.h>
# include<string.h>
# define maxn 1005        //这样定义max函数，是不是就不用考虑a，b的类型了
# define max(a,b) a>b?a:b
int dp[maxn][maxn];
char a[maxn],b[maxn];
int main(){
    int lena,lenb,i,j;
    while(gets(a)&&gets(b)){    //题目中这里不能用scanf，只能用gets输入
        lena=strlen(a);
        lenb=strlen(b);

        for(i=0;i<=lena;i++)
            for(j=0;j<=lenb;j++)
                dp[i][j] = 0;
            
            for(i=1;i<=lena;i++)
            {
                for(j=1;j<=lenb;j++)
                {
                    dp[i][j] = max(dp[i-1][j],dp[i][j-1]);    
                    if(a[i-1]==b[j-1])                        
                        dp[i][j] = max(dp[i][j] , dp[i-1][j-1] + 1);
                }            
            }    
            
            printf("%d\n",dp[lena][lenb]);
    }
    return 0;
}