字符串处理/贪心 Codeforces Round #307 (Div. 2) B. ZgukistringZ

 1 /*
 2     题意：任意排列第一个字符串，使得有最多的不覆盖a/b字符串出现
 3     字符串处理/贪心：暴力找到最大能不覆盖的a字符串，然后在b字符串中动态得出最优解
 4     恶心死我了，我最初想输出最多的a，再最多的b，然而并不能保证是最多的:(
 5 */
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <string>
 9 #include <iostream>
10 #include <algorithm>
11 #include <cmath>
12 using namespace std;
13 
14 const int MAXN = 1e5 + 10;
15 const int INF = 0x3f3f3f3f;
16 char s[MAXN], a[MAXN], b[MAXN];
17 int cnt_s[30], cnt_a[30], cnt_b[30];
18 int len_s, len_a, len_b;
19 
20 int main(void)        //Codeforces Round #307 (Div. 2) B. ZgukistringZ
21 {
22 //    freopen ("B.in", "r", stdin);
23 
24     while (scanf ("%s", s) == 1)
25     {
26         scanf ("%s", a);    scanf ("%s", b);
27         len_s = strlen (s);    len_a = strlen (a);    len_b = strlen (b);
28 
29         memset (cnt_s, 0, sizeof (cnt_s));
30         memset (cnt_a, 0, sizeof (cnt_a));
31         memset (cnt_b, 0, sizeof (cnt_b));
32 
33         for (int i=0; i<len_s; ++i)    cnt_s[s[i]-‘a‘]++;
34         for (int i=0; i<len_a; ++i)    cnt_a[a[i]-‘a‘]++;
35         for (int i=0; i<len_b; ++i)    cnt_b[b[i]-‘a‘]++;
36 
37         int ans_a = 0, ans_b = 0;    int tot = 100000;
38         for (int i=0; i<30; ++i)    {if (cnt_a[i])    tot = min (tot, cnt_s[i] / cnt_a[i]);}
39         for (int i=0; i<=tot; ++i)
40         {
41             int p = 100000;
42             for (int j=0; j<30; ++j)    {if (cnt_b[j])    p = min (p, (cnt_s[j]  - i * cnt_a[j]) / cnt_b[j]);}
43             if (i + p > ans_a + ans_b)    {ans_a = i;    ans_b = p;}
44         }
45 
46         for (int i=1; i<=ans_a; ++i)    printf ("%s", a);
47         for (int j=1; j<=ans_b; ++j)    printf ("%s", b);
48         for (int i=0; i<30; ++i)
49         {
50             for (int j=1; j<=cnt_s[i]-ans_a*cnt_a[i]-ans_b*cnt_b[i]; ++j)
51                 printf ("%c", ‘a‘ + i);
52         }
53         puts ("");
54     }
55 
56     return 0;
57 }