最长公共子序列

问题：求两个字符串的最长公共子序列
题目1042：Coincidence
时间限制：1 秒内存限制：32 兆特殊判题：否提交：4093解决：2241
题目描述：
Find a longest common subsequence of two strings.
输入：
First and second line of each input case contain two strings of lowercase character a…z. There are no spaces before, inside or after the strings. Lengths of strings do not exceed 100.
输出：
For each case, output k – the length of a longest common subsequence in one line.
样例输入：
abcd
cxbydz
样例输出：
2
来源：
2008年上海交通大学计算机研究生机试真题
分析：
递推公式

dp[0][j](0<=j<=m)dp[i][0](0<=i<=n)dp[i][j]=dp[i-1][j-1]+1;dp[i][j] =max(dp[i][j-1],dp[i-1][j]);

代码：

#include <iostream>#include <stdio.h>#include <string.h>#define MAX 101 /* run this program using the console pauser or add your own getch, system("pause") or input loop */int dp[MAX][MAX];int max(int a,int b){    if(a>b){        return a;    }else{        return b;    }}int main(int argc, char** argv) {    char str1[MAX],str2[MAX];    while(scanf("%s%s",str1,str2) != EOF){        int len1=strlen(str1);        int len2 = strlen(str2);        for(int i=0;i<len1;i++){            dp[i][0] = 0;        }        for(int i=0;i<len2;i++){            dp[0][i]= 0;        }        for(int i=1;i<=len1;i++){            for(int j=1;j<=len2;j++){                if(str1[i-1] == str2[j-1]){                    dp[i][j] = dp[i-1][j-1]+1;                }else{                    dp[i][j] = max(dp[i][j-1],dp[i-1][j]);                }            }        }        printf("%d\n",dp[len1][len2]);    }    return 0;}