【HDU2122】Ice_cream’s world III（MST基础题）

2坑，3次WA。

1.判断重边取小。2.自边舍去。

（个人因为vis数组忘记初始化，WA了3次，晕死！！）

 

 1 #include <iostream>

 2 #include <cstring>

 3 #include <cstdlib>

 4 #include <cstdio>

 5 #include <cmath>

 6 #include <cctype>

 7 #include <algorithm>

 8 #include <numeric>

 9 

10 #define typec int

11 using namespace std;

12 

13 const int V = 1005;

14 const typec inf = 0xffff;

15 int vis[V];

16 typec lowc[V];

17 int Map[V][V];

18 

19 typec prim (typec cost[][V], int n) {

20     int i, j, p;

21     typec minc, res = 0;

22     memset(vis, 0, sizeof(vis));

23     vis[0] = 1;

24     for (i = 1; i < n; ++ i) lowc[i] = cost[0][i];

25     for (i = 1; i < n; ++ i) {

26         minc = inf; p = -1;

27         for (j = 0 ; j < n; ++ j) {

28             if (0 == vis[j] && minc > lowc[j]) {

29                 minc = lowc[j]; p = j;

30             }

31         }

32         if (inf == minc) return -1;

33         res += minc; vis[p] = 1;

34         for (j = 0 ; j < n; ++ j) {

35             if (0 == vis[j] && lowc[j] > cost[p][j]) {

36                 lowc[j] = cost[p][j];

37             }

38         }

39     }

40     return res;

41 }

42 

43 int main () {

44     int n, road_n;

45     int cur = 0;

46     while (cin >> n >> road_n) {

47         /*if (cur ++ != 0) {

48             cout << endl;

49         }*/

50         for (int i = 0 ; i < n; ++ i) {

51             for (int j = 0 ; j < n; ++ j) {

52                 if (i == j) Map[i][j] = 0;

53                 else {

54                     Map[i][j] = inf;

55                 }

56             }

57         }

58 

59         for (int i = 0 ; i < road_n; ++ i) {

60             int x, y, c;

61             scanf("%d%d%d", &x, &y, &c);

62             if (x == y) {

63                 continue;

64             }

65             Map[x][y] = Map[y][x] = min(Map[x][y], c);

66         }

67 

68         int ans = prim (Map, n);

69 

70         if (ans == -1) {

71             cout << "impossible" << endl << endl;

72         } else {

73             cout << ans << endl << endl;

74         }

75     }

76     return 0;

77 }