Aizu0189 Convenient Location（多源最短路）

Posted 2020-11-02 Surprisez

https://vjudge.net/problem/Aizu-0189

题意:求某一点到其他所有点的最短路径之和，输出该点与和。

注意Floyd可以求多源最短路径，而Dijkstra只能求单源。

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<iostream>
 4 #include<algorithm> 
 5 #include<cstring>
 6 #include<vector>
 7 #include<map>
 8 #include<set>
 9 #define LL long long
10 #define maxn 100005
11 #define MOD 1e9+7
12 #define INF 0x3f3f3f3f
13 typedef long long ll;
14 using namespace std;
15 int n, D[110][110], x, y, z, maxm, mini;
16 int main()
17 {
18     while(cin >> n, n){
19         maxm = -INF;
20         for(int i = 0; i < 10; i++){
21             for(int j = 0; j < 10; j++){
22                 D[i][j] = INF;
23             }
24         }
25         for(int i = 0; i < n; i++){
26             cin >> x >> y >> z;
27             D[x][y] = min(D[x][y], z); 
28             D[y][x] = D[x][y];
29             maxm = max(maxm, max(x, y));
30         }
31         maxm++;
32         for(int i = 0; i < 10; i++){
33             D[i][i] = 0;
34         }
35         for(int k = 0; k < maxm; k++){
36             for(int i = 0; i < maxm; i++){
37                 for(int j = 0; j < maxm; j++){
38                     if(D[i][j] > D[i][k]+D[k][j]){
39                         D[i][j] = D[i][k]+D[k][j];
40                     }
41                 }
42             }
43         }
44         mini = INF;
45         int loc = -1;
46         for(int i = 0; i < maxm; i++){
47             int sum = 0;
48             for(int j = 0; j < maxm; j++){
49                 sum += D[i][j];
50             }
51             if(mini > sum){
52                 mini = sum;
53                 loc = i;
54             } 
55         }
56         cout << loc << " " << mini << endl;
57     }
58     return 0;
59 }