UVa 10252 Common Permutation (water ver.)

10252 - Common Permutation

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1193

Given two strings of lowercase letters, a and b, print the longest string x of lowercase letters such that there is a permutation of x that is a subsequence of a and there is a permutation ofx that is a subsequence of b.

Input

Input file contains several lines of input. Consecutive two lines make a set of input. That means in the input file line1 and 2 is a set of input, line 3 and4 is a set of input and so on. The first line of a pair containsa and the second contains b. Each string is on a separate line and consists of at most1000 lowercase letters.

Output

For each set of input, output a line containingx. If several x satisfy the criteria above, choose the first one in alphabetical order.

Sample Input:

pretty

women

walking

down

the

street

Sample Output:

e

nw

et

水。

完整代码：

/*0.019s*/

#include<bits/stdc++.h>
using namespace std;  

char a[1005], b[1005], ans[1005];
bool vis[1005];  

void solve(char* a, char* b)
{
    int lena = strlen(a), lenb = strlen(b), i, j, c = 0;
    memset(vis, 0, sizeof(vis));
    for (i = 0; i < lena; ++i)
        for (j = 0; j < lenb; ++j)
        {
            if (!vis[j] && a[i] == b[j])
            {
                ans[c++] = a[i];
                vis[j] = true;
                break;
            }
        }
    ans[c] = 0;
    sort(ans, ans + c);
    puts(ans);
}  

int main()
{
    while (gets(a))
    {
        gets(b);
        if (strlen(a) < strlen(b)) solve(a, b);
        else solve(b, a);
    }
    return 0;
}

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