HDU1159——Common Subsequence

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

分析：用一个二维的表格来记录以前的最优值，递推关系式为：

当前字符str1[i] == str2[j]，dp[i][j] = dp[i -1][j-1] + 1；

否则，dp[i][j] = max(dp[i -1][j] ,dp[i][j -1]);

参考代码：

#include<stdio.h>#include<string.h>#define M 2010char str1[M];char str2[M];int dp[M][M];inline int max(const int a, const int b){return a > b ? a : b;}void main(){int nLen1, nLen2;int i,j;int nMax;while(scanf("%s %s",&str1,&str2) != EOF){nLen1 = strlen(str1);nLen2 = strlen(str2);memset(dp,0,sizeof(dp));nMax = 0;for(i = 1; i <= nLen1; ++i){for(j = 1; j <= nLen2; ++j){if(str1[i - 1] == str2[j - 1]){dp[i][j] = dp[i - 1][j - 1] + 1;}else{dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);}nMax = max(nMax, dp[i][j]);}}printf("%d\n",nMax);}}