1110 - An Easy LCS

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LCS means ‘Longest Common Subsequence‘ that means two non-empty strings are given; you have to find the Longest Common Subsequence between them. Since there can be many solutions, you have to print the one which is the lexicographically smallest. Lexicographical
order means dictionary order. For example, ‘abc‘ comes before ‘abd‘ but ‘aaz‘ comes before ‘abc‘.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a blank line. The next two lines will contain two strings of length 1 to 100. The strings contain lowercase English characters only.

Output

For each case, print the case number and the lexicographically smallest LCS. If the LCS length is 0 then just print ‘:(‘.

题意：找出最长公共自序列，有一样长的取字典序小的。如果lcs长度为0，输出哭脸。

做法：在最长公共子序列基础上，每次dp的时候 判断下两个string的字典序大小。最长公共子序列的复杂度是100*100，字符串最长是100，O（10^6）。

#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
using namespace std;
#define MAX_N 110//再大dp数组会爆的   O(n*m)
char a[MAX_N],b[MAX_N];
int dp[MAX_N+1][MAX_N+1];
string dps[MAX_N][MAX_N];
int mxxx(int a,int b)
{
	return a>b?a:b;
}
int main()
{
	int n,m,i,j,tem;
	int t;
	int cas=1;
	scanf("%d",&t);
	while(t--)//a b链的长度
	{
		memset(dp,0,sizeof(dp));//dp中ij 代表a链前i项  和b链 前j项  最大公共子序列长度
		scanf("%s",a);
		scanf("%s",b);
		n=strlen(a);
		m=strlen(b);

		for(i=0;i<=n;i++)
			for(int j=0;j<=m;j++)
				dps[i][j].clear();
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{
				if(a[i]==b[j])//dp的ij 从1开始算
				{ 

					if(dp[i+1][j]>dp[i][j+1])
						dps[i+1][j+1]=dps[i+1][j];
					else if(dp[i+1][j]<dp[i][j+1])
						dps[i+1][j+1]=dps[i][j+1];
					else if(dps[i+1][j]<dps[i][j+1])
						dps[i+1][j+1]=dps[i+1][j];
					else
						dps[i+1][j+1]=dps[i][j+1];
					dp[i+1][j+1]=mxxx(dp[i+1][j],dp[i][j+1]);

					if(dp[i+1][j+1]>dp[i][j]+1)
						dps[i+1][j+1]=dps[i+1][j+1];
					else if(dp[i+1][j+1]<dp[i][j]+1)
						dps[i+1][j+1]=dps[i][j]+a[i];
					else if(dps[i+1][j+1]>dps[i][j]+a[i])
						dps[i+1][j+1]=dps[i][j]+a[i];
					else
						dps[i+1][j+1]=dps[i+1][j+1]; 

					dp[i+1][j+1]=mxxx(dp[i][j]+1,mxxx(dp[i+1][j],dp[i][j+1]));
				}
				else
				{
					if(dp[i+1][j]>dp[i][j+1])
						dps[i+1][j+1]=dps[i+1][j];
					else if(dp[i+1][j]<dp[i][j+1])
						dps[i+1][j+1]=dps[i][j+1];
					else if(dps[i+1][j]<dps[i][j+1])
						dps[i+1][j+1]=dps[i+1][j];
					else
						dps[i+1][j+1]=dps[i][j+1];
					dp[i+1][j+1]=mxxx(dp[i+1][j],dp[i][j+1]);
				}
			}
		}
		printf("Case %d: ",cas++);
		if(dp[n][m])
		cout<<dps[n][m]<<endl;
		else
			puts(":(");
	}
	return 0;
}