UVa 140 带宽

题意：给出一个n个结点的图G和一个结点的排列，定义结点的带宽为i和相邻结点在排列中的最远距离，求出让带宽最小的结点排列。

思路:用STL的next_permutation来做确实是很方便，适当剪枝一下就可以了，不过我不明白的是为什么我用string字符串会超时...

 #include<iostream>
 #include<cstring>
 #include<string>
 #include<algorithm>
 using namespace std;

 int a[], c[];
 int ans[];
 int map[][];
 char str[];

 int main()
 {
     //freopen("D:\\txt.txt", "r", stdin);
     int l, k, maxn,n;
     while (gets(str) )
     {
         if (str[] == '#') break;
         memset(map, , sizeof(map));
         memset(a, , sizeof(a));
         memset(ans, , sizeof(ans));
         memset(c, , sizeof(c));
         l = strlen(str);         for (int i = ; i < l; i++)
         {
             if (str[i] == ':')
             {
                 int t = str[i - ] - 'A';
                 a[t] = ;
                 for (k = i + ; str[k] != ';' && k < l; k++)
                 {
                     int p = str[k] - 'A';
                     a[p] = ;
                     map[t][p] = map[p][t] = ;
                 }
                 i = k;
             }
         }

         n = ;
         for (int i = ; i < ; i++)
         {
             if (a[i])
                 c[n++] = i;
         }
         maxn = ;
         do
         {
             int sum = ;
             int ok = ;
             for (int i = ; i < n; i++)
             {
                 for (int j = i + ; j < n; j++)
                 {
                     if (map[c[i]][c[j]])
                     if (j - i>sum)
                         sum = j - i;
                     if (sum>=maxn)  { ok = ; break; }
                 }
                 if (!ok)  break;
             }

             if (sum < maxn)
             {
                 maxn = sum;
                 memcpy(ans, c, sizeof(c));
             }
         } while (next_permutation(c, c + n));
         for (int i = ; i < n; i++)
         {
             char s = 'A' + ans[i];
             cout << s << " ";
         }
         cout << "-> " << maxn << endl;
     }
     return ;
 }