贪心 UVALive 6832 Bit String Reordering

题目传送门

 1 /*
 2     贪心：按照0或1开头，若不符合，选择后面最近的进行交换。然后选取最少的交换次数
 3 */
 4 #include <cstdio>
 5 #include <algorithm>
 6 #include <cstring>
 7 #include <string>
 8 #include <cmath>
 9 #include <vector>
10 #include <map>
11 #include <queue>
12 using namespace std;
13
14 const int MAXN = 22 + 10;
15 const int INF = 0x3f3f3f3f;
16 int a[MAXN], b[MAXN], c[MAXN];
17
18 int main(void)        //UVALive 6832 Bit String Reordering
19 {
20 //    freopen ("A.in", "r", stdin);
21
22     int n, m;
23     while (scanf ("%d%d", &n, &m) == 2)
24     {
25         for (int i=1; i<=n; ++i)    {scanf ("%d", &a[i]);    c[i] = a[i];}
26         for (int i=1; i<=m; ++i)    scanf ("%d", &b[i]);
27
28         int cnt1 = 0, cnt2 = 0;    int now = 0;    int p = 0;
29         bool ok1 = true, ok2 = true;
30         for (int i=1; i<=m && ok1; ++i)
31         {
32             for (int j=1; j<=b[i]; ++j)
33             {
34                 if (a[p+j] != now)
35                 {
36                     int k = p + j;
37                     while (k <= n && a[k] != now)    k++;
38                     if (k == n+1 || a[k] != now)    {ok1 = false;    break;}
39                     cnt1 += k - (p + j);
40                     swap (a[k], a[p+j]);
41                 }
42             }
43             p += b[i];    now = 1 - now;
44         }
45
46         now = 1;    p = 0;
47         for (int i=1; i<=m && ok2; ++i)
48         {
49             for (int j=1; j<=b[i]; ++j)
50             {
51                 if (c[p+j] != now)
52                 {
53                     int k = p + j;
54                     while (k <= n && c[k] != now)    k++;
55                     if (k == n+1 || c[k] != now)    {ok2 = false;    break;}
56                     cnt2 += k - (p + j);
57                     swap (c[p+j], c[k]);
58                 }
59             }
60             p += b[i];    now = 1 - now;
61         }
62
63 //        printf ("%d %d\n", cnt1, cnt2);
64         if (!ok1)    printf ("%d\n", cnt2);
65         else if (!ok2)    printf ("%d\n", cnt1);
66         else    printf ("%d\n", min (cnt1, cnt2));
67     }
68
69     return 0;
70 }