暑假集训第四周阶段二DP A - 最长公共子序列

A - Common Subsequence

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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 abfcabprogramming contest abcd mnp 

 

Sample Output

420 

分析：

动态规划的题一般都有一定的思想，也有一些比较典型，比较经典的题目

这就是典型的最长公共子序列，两个for循环，一直保持最大值，到最后找到最优解

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#include<stdio.h>#include<string.h> char s1[1005],s2[1005]; int dp[1005][1005];int main(){    int len1,len2,i,j;    while(scanf("%s %s",s1,s2)!=EOF)    {        len1=strlen(s1),len2=strlen(s2);        memset(dp,0,sizeof(dp));        for(i=1;i<=len1;i++)            for(j=1;j<=len2;j++)            if(s1[i-1]==s2[j-1])            dp[i][j]=dp[i-1][j-1]+1;        else            dp[i][j]=dp[i-1][j]>dp[i][j-1]?dp[i-1][j]:dp[i][j-1];        printf("%d\n",dp[len1][len2]);    }    return 0;}