最短路(Dijkstra) POJ 1062 昂贵的聘礼

 

题目传送门

 1 /*  2  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  3  更新每个点的最短路的最小值  4  注意：国王的等级不一定是最高的:)  5 */  6 #include <cstdio>  7 #include <iostream>  8 #include <algorithm>  9 #include <cmath> 10 #include <map> 11 #include <vector> 12 #include <cstring> 13 #include <string> 14 using namespace std; 15 16 const int MAXN = 1e2 + 10; 17 const int INF = 0x3f3f3f3f; 18 19 int d[MAXN]; 20 int cost[MAXN][MAXN]; 21 int x[MAXN]; 22 int lv[MAXN]; 23 int used[MAXN]; 24 25 int Dijkstra(int n) 26 { 27 for (int i=1; i<=n; ++i) d[i] = cost[0][i]; 28 29 for (int i=1; i<=n; ++i) 30  { 31 int x = 0; 32 int mn = INF; 33 for (int j=1; j<=n; ++j) 34  { 35 if (!used[j] && d[j] < mn) mn = d[x=j]; 36  } 37 if (x == 0) break; 38 used[x] = 1; 39 for (int j=1; j<=n; ++j) 40  { 41 if (!used[j] && cost[x][j] > 0) 42 d[j] = min (d[j], d[x] + cost[x][j]); 43  } 44  } 45 46 return d[1]; 47 } 48 49 int main(void) //POJ 1062 昂贵的聘礼 50 { 51 //freopen ("C.in", "r", stdin); 52 53 int n, m; 54 55 while (~scanf ("%d%d", &m, &n)) 56  { 57 memset (used, 0, sizeof (used)); 58 memset (d, 0, sizeof (d)); 59 memset (cost, 0, sizeof (cost)); 60 for (int i=1; i<=n; ++i) 61  { 62 scanf ("%d%d%d", &cost[0][i], &lv[i], &x[i]); 63 for (int j=1; j<=x[i]; ++j) 64  { 65 int t, u; 66 scanf ("%d%d", &t, &u); 67 cost[t][i] = u; 68  } 69  } 70 71 int tmp, ans = INF, maxlv; 72 for (int i=1; i<=n; ++i) 73  { 74 maxlv = lv[i]; 75 for (int j=1; j<=n; ++j) 76  { 77 if (lv[j] > maxlv || maxlv - lv[j] > m) used[j] = 1; 78 else used[j] = 0; 79  } 80 tmp = Dijkstra (n); 81 ans = min (ans, tmp); 82  } 83 84 printf ("%d\n", ans); 85  } 86 87 return 0; 88 }