POJ -- 1458 Common Subsequence

最新推荐文章于 2019-07-05 08:49:53 发布

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Common Subsequence

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu

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

DP思想，LIS（公共子序列的最大长度）。dp[i][j]表示序列a的前i项和序列b的前j项(不包括a[i][j]项）的最大子序列的长度。举个栗子：

A序列：a c d b d

B序列：b c a d b

i=1:dp[1][1]=0，dp[1][2]=0,dp[1][3]=1,dp[1][4]=1,dp[1][5]=1;

i=2:dp[2][1]=0，dp[2][2]=1,dp[2][3]=1,dp[2][4]=1,dp[2][5]=1;

i=3:dp[3][1]=0，dp[3][2]=1,dp[3][3]=1,dp[3][4]=2,dp[1][5]=2;

i=4:dp[4][1]=1，dp[4][2]=1,dp[4][3]=1,dp[4][4]=2,dp[4][5]=3;

i=5:dp[5][1]=1，dp[5][2]=1,dp[5][3]=1,dp[5][4]=2,dp[5][5]=3;

这个过程类似于找朋友，两个序列的每一项都要比一比！！

于是可以得出状态转移方程:

如果a[i-1] == b[j-1]，dp[i][j] = dp[i-1][j-1] + 1;

如果a[i-1] != b[j-1]，dp[i][j] = max(dp[i-1][j],dp[i][j-1])

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char a[1005],b[1005];
int dp[1005][1005];
int main(){
	int la,lb;
	while(~scanf("%s%s",&a,&b)){
		la=strlen(a);
		lb=strlen(b);
		memset(dp,0,sizeof(dp));
		for(int i=1;i<=la;i++){//从1开始防止数组越界 
			for(int j=1;j<=lb;j++){
				if(a[i-1]==b[j-1])
					dp[i][j]=dp[i-1][j-1]+1;
				else
					dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
			}
		}
		printf("%d\n",dp[la][lb]);
	}
	return 0;
}