uva140 STL枚举排列加剪枝

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn = 10;
const int INF = 0x7fffffff;

int main()
{
    char s[1000];
    while( scanf("%s", s) && s[0]!= '#')              
    {
        int n = 0;
        int id[300] = {};
        char letter[maxn] = {};
        for(char c= 'A'; c <= 'Z'; c++)
            if(strchr(s, c) != NULL)
            {
                id[c] = n++;
                letter[id[c]] = c;                        //建立双射数组 id: char - int  letter: int - char 
            }
        int p = 0, q = 0, len = strlen(s);
        
        vector<int> u, v;
        for(;;)  //使用 p q 定位 ';' 与 ';' 
        {
            while(s[p] != ':' && p < len)
              p++;
            if( p == len)                                //p位于q前 到底退出 
              break;
            while(s[q] != ';' && q < len)
              q++;
            for(int i = p+1; i < q; i++)
            {
                u.push_back(id[s[p-1]]);
                v.push_back(id[s[i]]);
            }
            p++;
            q++;                                          //继续移动 
        }
//      for(int i = 0; i < u.size(); i++)
//      {
//          cout<<u[i]<<" "<<v[i]<<endl;    
//      }  
        
        int Pos[maxn] = {}, P[maxn] = {}, bestP[maxn] = {};    //P[i] = j j放置在i上   Pos[i] = j i在位置j上 
        int bandwidth = 0, ans = INF;
        for(int i = 0; i < n; i++)
            P[i] = i;
        do{
            for(int i = 0; i < n; i++)
                Pos[P[i]] = i;
            bandwidth = 0;
            for(int i = 0; i < u.size(); i++)
            {
                bandwidth = max(bandwidth, abs(Pos[u[i]] - Pos[v[i]]));
                if(bandwidth > ans)
                    break;
            }
            if(bandwidth < ans)
            {
                ans = bandwidth;
                memcpy(bestP, P, sizeof(P));
            }
        }while(next_permutation(P, P+n));
        for(int i = 0; i < n; i++)
            printf("%c ", letter[bestP[i]]);
        printf("-> %d\n", ans);
    }
    return 0;
}