HDU 1159

最新推荐文章于 2019-08-21 02:59:11 发布

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最新推荐文章于 2019-08-21 02:59:11 发布

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本文链接：https://blog.youkuaiyun.com/asd7788999/article/details/80299286

本文介绍了一种解决最长公共子序列问题的算法实现。通过动态规划方法，该程序能够接收两组字符串作为输入，并输出这两组字符串之间的最长公共子序列的长度。文章提供了完整的C++代码示例。

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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

最长公共字串，没什么要仔细讲的东西，直接输入两个字符串就行了，不用在意空格，字符数组开1005就够用了

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;

int dp[1005][1005];
char a[1005];
char b[1005];
int main(){
	int t;
	while(~scanf("%s",&a[1])){
		int i,j;
		scanf("%s",&b[1]);
		int m,n;
		a[0]='0';
		b[0]='0';
		n=strlen(a);
		m=strlen(b);
//		printf("--%c--%c--\n",a[1],b[1]);
		memset(dp,0,sizeof(dp));
		for(i=1;i<strlen(a);i++){
			for(j=1;j<strlen(b);j++){
				if(a[i]==b[j]) dp[i][j]=dp[i-1][j-1]+1;
				else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
//				printf("!%d %d %d\n",i,j,dp[i][j]);
			}
		}
		printf("%d\n",dp[n-1][m-1]);
	}
}