最短路(Dijkstra) POJ 1062 昂贵的聘礼

题目传送门

 /*
     最短路：Dijkstra算法，首先依照等级差距枚举“删除”某些点，即used，然后分别从该点出发生成最短路
             更新每个点的最短路的最小值
     注意：国王的等级不一定是最高的:)
 */
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cmath>
 #include <map>
 #include <vector>
 #include <cstring>
 #include <string>
 using namespace std;
 const int MAXN = 1e2 + ;
 const int INF = 0x3f3f3f3f;
 int d[MAXN];
 int cost[MAXN][MAXN];
 int x[MAXN];
 int lv[MAXN];
 int used[MAXN];
 int Dijkstra(int n)
 {
     for (int i=; i<=n; ++i)    d[i] = cost[][i];
     for (int i=; i<=n; ++i)
     {
         int x = ;
         int mn = INF;
         for (int j=; j<=n; ++j)
         {
             if (!used[j] && d[j] < mn)        mn = d[x=j];
         }
         if (x == )        break;
         used[x] = ;
         for (int j=; j<=n; ++j)
         {
             if (!used[j] && cost[x][j] > )
                 d[j] = min (d[j], d[x] + cost[x][j]);
         }
     }
     return d[];
 }
 int main(void)        //POJ 1062 昂贵的聘礼
 {
     //freopen ("C.in", "r", stdin);
     int n, m;
     while (~scanf ("%d%d", &m, &n))
     {
         memset (used, , sizeof (used));
         memset (d, , sizeof (d));
         memset (cost, , sizeof (cost));
         for (int i=; i<=n; ++i)
         {
             scanf ("%d%d%d", &cost[][i], &lv[i], &x[i]);
             for (int j=; j<=x[i]; ++j)
             {
                 int t, u;
                 scanf ("%d%d", &t, &u);
                 cost[t][i] = u;
             }
         }
         int tmp, ans = INF, maxlv;
         for (int i=; i<=n; ++i)
         {
             maxlv = lv[i];
             for (int j=; j<=n; ++j)
             {
                 if (lv[j] > maxlv || maxlv - lv[j] > m)        used[j] = ;
                 else        used[j] = ;
             }
             tmp = Dijkstra (n);
             ans = min (ans, tmp);
         }
         printf ("%d\n", ans);
     }
     return ;
 }