poj3268 Silver Cow Party（两次dijkstra）

Posted 2020-11-03 Surprisez

https://vjudge.net/problem/POJ-3268

一开始floyd超时了。。

对正图定点求最短，对逆图定点求最短，得到任意点到定点的往返最短路。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<stack>
 8 #define lson l, m, rt<<1
 9 #define rson m+1, r, rt<<1|1
10 #define INF 0x3f3f3f3f
11 typedef unsigned long long ll;
12 using namespace std;
13 int t[1010][1010], rt[1010][1010];
14 int d[1010], rd[1010], vis[1010];
15 int n, m, x, a, b, T;
16 void dijkstra(int dist[], int a[][1010])
17 {
18     memset(vis, 0, sizeof(vis));
19     for(int i = 1; i <= n; i++){
20         dist[i] = INF;
21     }
22     dist[x] = 0;
23     for(int i = 1; i <= n; i++){
24         int mini = INF, k = -1;
25         for(int j = 1; j <= n; j++){
26             if(!vis[j]&&mini > dist[j]){
27                 mini = dist[j];
28                 k = j;
29             }
30         }
31         vis[k] = 1;
32         for(int j = 1; j <= n; j++){
33             if(!vis[j]&&dist[j] > dist[k]+a[k][j]){
34                 dist[j] = dist[k]+a[k][j];
35             }
36         }
37     }
38 }
39 int main()
40 {
41     cin >> n >> m >> x;
42     for(int i = 1; i <= n; i++){
43         for(int j = 1; j <= n; j++){
44             t[i][j] = INF; rt[i][j] = INF;
45         }
46     }
47     for(int i = 1; i <= m; i++){
48         cin >> a >> b >> T;
49         t[a][b] = T;//图 
50         rt[b][a] = T;//逆图 
51     }
52     dijkstra(d, t);
53     dijkstra(rd, rt);
54     int maxm = -INF;
55     for(int i = 1; i <= n; i++){
56         if(d[i]!=INF&&rd[i]!=INF&&d[i]+rd[i] > maxm){
57             maxm = d[i]+rd[i];
58         }
59     }
60     cout << maxm << endl;
61     return 0;
62 }