题目传送门
1 /*
2 最短路：Floyd模板题
3 主要是两点最短的距离和起始位置
4 http://blog.csdn.net/y990041769/article/details/37955253
5 */
6 #include <cstdio>
7 #include <iostream>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <string>
12 #include <vector>
13 using namespace std;
14
15 const int MAXN = 100 + 10;
16 const int INF = 0x3f3f3f3f;
17 int d[MAXN][MAXN];
18
19 void Floyd_Warshall(int n)
20 {
21 for (int k=1; k<=n; ++k)
22 {
23 for (int i=1; i<=n; ++i)
24 {
25 for (int j=1; j<=n; ++j)
26 {
27 if (d[i][j] > d[i][k] + d[k][j])
28 d[i][j] = d[i][k] + d[k][j];
29 }
30 }
31 }
32 }
33
34 void work(int n)
35 {
36 Floyd_Warshall (n);
37 int res; int ans = INF; int num;
38 for (int i=1; i<=n; ++i)
39 {
40 res = 0;
41 for (int j=1; j<=n; ++j)
42 {
43 if (i == j) continue;
44 if (res < d[i][j]) res = d[i][j];
45 }
46 if (ans > res)
47 {
48 ans = res;
49 num = i;
50 }
51 }
52
53 printf ("%d %d
", num, ans);
54 }
55
56 int main(void) //POJ 1125 Stockbroker Grapevine
57 {
58 //freopen ("E.in", "r", stdin);
59
60 int n, m;
61 while (~scanf ("%d", &n) && n)
62 {
63 for (int i=1; i<=n; ++i)
64 for (int j=1; j<=n; ++j) d[i][j] = INF;
65 for (int i=1; i<=n; ++i)
66 {
67 scanf ("%d", &m);
68 int x, y;
69 for (int j=1; j<=m; ++j)
70 {
71 scanf ("%d%d", &x, &y);
72 d[i][x] = y;
73 }
74 }
75
76 work (n);
77 }
78
79 return 0;
80 }