POJ 1458 Common Subsequence DP

Common Subsequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 54786 Accepted: 22832

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, x_ij = 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.

Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

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

Source

Southeastern Europe 2003

在这道题中递推关系为

当前字母相同，在原有长度加1

当前字母不同，在两者的长度中选一个大的

问题的难点在于DP数组的设定，因为是两个字符串，就要开一个二维数组DP[i][j], 用i代表第一个字符串中位置，用j代表在第二个字符串中位置

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

还可以在这个过程中记录所走的路程并输出相同字符串,，可我真的不会写啊