POJ - 1458 Common Subsequence

最新推荐文章于 2019-07-05 08:49:53 发布

原创最新推荐文章于 2019-07-05 08:49:53 发布 · 269 阅读

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Common Subsequence

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

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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 _{i_j} = 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         abfcab
programming    contest 
abcd           mnp

Sample Output

4
2
0

Source

Southeastern Europe 2003

DP求最长公共子序列的经典题，基础必会题。代码：

//这道题没有给出字符串的大小，测了一下，只要大于200就可以。

#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
using namespace std;
const int maxn=1000;
char x[maxn],y[maxn];
int dp[maxn][maxn];
int main(){
int i,j;
while(scanf("%s%s",x,y)!=EOF)
{
int a=strlen(x);
int b=strlen(y);
memset(dp,0,sizeof(dp));
for(i=1;i<=a;i++)
for(j=1;j<=b;j++)
{
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);//dp的关键步骤一般就几步，本题最关键的是以下三句。但都较好理解
if(x[i-1]==y[j-1])
dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1);
}
printf("%d\n",dp[a][b]);
}
return 0;
}