HDU 1159 Common Subsequence （dp）

题目链接

Problem 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 abfcab
programming contest
abcd mnp

Sample Output

4
2
0

分析：

给出两个序列S1和S2，求出的这两个序列的最大公共部分S3就是就是S1和S2的最长公共子序列了。公共部分

必须是以相同的顺序出现，但是不必要是连续的。

1、字符相同，则指向左上，并加1

2、字符不同，则指向左边或者上边较大的那个

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
char ch1[1000],ch2[1000];
int dp[1000][1000];
int k1,k2;
void zhixul()
{
    int i,j;
    memset(dp,0,sizeof(dp));
    for(i = 1; i<=k1; i++)
    {
        for(j = 1; j<=k2; j++)
        {
            if(ch1[i-1] == ch2[j-1])///字符相同，则指向左上，并加1
                dp[i][j] = dp[i-1][j-1]+1;
            else
                dp[i][j] = max(dp[i-1][j],dp[i][j-1]);///字符不同，比较左边和上边那个大取最大
        }
    }
}
int main()
{
    while(~scanf("%s%s",ch1,ch2))
    {
        k1 = strlen(ch1);
        k2 = strlen(ch2);
        zhixul();
        printf("%d\n",dp[k1][k2]);
    }
    return 0;
}