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[30], c[30];
int ans[30];
int map[30][30];
char str[500];

int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    int l, k, maxn,n;
    while (gets(str) )
    {
        if (str[0] == '#') break;
        memset(map, 0, sizeof(map));
        memset(a, 0, sizeof(a));
        memset(ans, 0, sizeof(ans));
        memset(c, 0, sizeof(c));
        l = strlen(str);25         for (int i = 0; i < l; i++)
        {
            if (str[i] == ':')
            {
                int t = str[i - 1] - 'A';
                a[t] = 1;
                for (k = i + 1; str[k] != ';' && k < l; k++)
                {
                    int p = str[k] - 'A';
                    a[p] = 1;
                    map[t][p] = map[p][t] = 1;
                }
                i = k;
            }
        }


        n = 0;
        for (int i = 0; i < 26; i++)
        {
            if (a[i])
                c[n++] = i;
        }
        maxn = 10000;
        do
        {
            int sum = 0;
            int ok = 1;
            for (int i = 0; i < n; i++)
            {
                for (int j = i + 1; j < n; j++)
                {
                    if (map[c[i]][c[j]])
                    if (j - i>sum)
                        sum = j - i;
                    if (sum>=maxn)  { ok = 0; break; }
                }
                if (!ok)  break;
            }
            
            if (sum < maxn)
            {
                maxn = sum;
                memcpy(ans, c, sizeof(c));
            }
        } while (next_permutation(c, c + n));
        for (int i = 0; i < n; i++)
        {
            char s = 'A' + ans[i];
            cout << s << " ";
        }
        cout << "-> " << maxn << endl;
    }
    return 0;
}