program ... clas mission ems 字符 mem example div

題目鏈接：

http://acm.hdu.edu.cn/showproblem.php?pid=1159

題目類型：

最大公共子串

題意概括：

給出兩個字符串，求這兩個字符串的最大公共子串

解題思路：

通過一個二維字符串，將兩個字符串進行比較，遇到相同則將左上角的值+1賦予當前位置的值，如果不相同，則將左邊和上面的值的最大值賦予當前值，右下角的值即最大公共子串的長度。

題目：

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38605 Accepted Submission(s): 17735

Problem Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <x1, x2, ..., xm> another sequence Z = <z1, z2, ..., zk> is a subsequence of X if there exists a strictly increasing sequence <i1, i2, ..., ik> of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = <a, b, f, c> is a subsequence of X = <a, b, c, f, b, c> with index sequence <1, 2, 4, 6>. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.

Sample Input abcfbc abfcab programming contest abcd mnp

Sample Output 4 2 0

# include <stdio.h>
# include <string.h>
int dp[1010][1010];
int maxx(int a,int b)
{
    if(a>b)
        return a;
    else
        return b;
}


int main ()
{
    int i,j,l1,l2;
    
    char a[1010],b[1010];
    
    while(scanf("%s%s",a+1,b+1)!=EOF)
    {
        int l1=strlen(a);
        int l2=strlen(b);
        memset(dp,0,sizeof(dp));
        for(i=0;i<l1;i++)
        {
            for(j=0;j<l2;j++)
            {
                if(i==0 || j==0)
                    dp[i][j]=0;
                else if(a[i]==b[j])
                    dp[i][j]=dp[i-1][j-1]+1;
                else if(a[i]!=b[j])
                    dp[i][j]=maxx(dp[i-1][j],dp[i][j-1]);
            }
        }
        printf("%d\n",dp[l1-1][l2-1]);
    }
}

hdu-1159 Common Subsequence