uestc 方老师分身 I

题意有点晕啊。。 水题一道。。

分身去教室的时候是单向的，回来的时候也是单向的，这时候就要反过来跑一遍最短路了。。

Dijkstra看着模板写的。。sigh~ 要不是渣渣。。还是记不住。。

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 #include<string>

 6 #include<queue>

 7 #include<algorithm>

 8 #include<map>

 9 #include<iomanip>

10 #include<climits>

11 #include<string.h>

12 #include<cmath>

13 #include<stdlib.h>

14 #include<vector>

15 #define INF 1e7

16 #define MAXN 100010

17 #define maxn 1000010

18 #define Mod 1000007

19 #define N 1010

20 using namespace std;

21 typedef long long LL;

22 

23 int n, m, x;

24 int G[N][N];

25 int d[N], d1[N], d2[N];

26 bool vis[N];

27 

28 bool Dijkstra(int x, int* d)

29 {

30     int i, j, pos, min;

31     for (i = 1; i <= n; ++i)

32         d[i] = G[x][i];

33     d[x] = 0;

34     vis[x] = true;

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

36         min = INF;

37         for (j = 1; j <= n; ++j) {

38             if (d[j] < min && !vis[j]) {

39                 pos = j;

40                 min = d[j];

41             }

42         }

43         if (min == INF) return false;

44         vis[pos] = true;

45         for (j = 1; j <= n; ++j)

46             if (d[pos] + G[pos][j] < d[j] && !vis[j])

47                 d[j] = d[pos] + G[pos][j];

48     }

49     return true;

50 }

51 

52 void process()

53 {

54     int u, v, t;

55     for (int i = 0; i <= n; ++i)

56         for (int j = 0; j <= n; ++j)

57             G[i][j] = INF;

58     for (int i = 0; i < m; ++i) {

59         cin >> u >> v >> t;

60         G[u][v] = t;

61     }

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

63     Dijkstra(x, d1);

64     for (int i = 0; i <= n; ++i)

65         for (int j = i + 1; j <= n; ++j)

66             swap(G[i][j], G[j][i]);

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

68     Dijkstra(x, d2);

69     for (int i = 0; i <= n; ++i)

70         d[i] = d1[i] + d2[i];

71     /*for (int i = 1; i <= n; ++i)

72         cout << d[i] << " ";

73     cout << endl;*/

74     int ans = 0;

75     for (int i = 1; i <= n; ++i)

76             ans = max(ans, d[i]);

77     cout << ans << endl;

78 }

79 

80 int main()

81 {

82     while (cin >> n >> m >> x)

83         process();

84     return 0;

85 }