POJ 3268 Silver Cow Party

Posted 2020-08-31 Lorazepam

http://poj.org/problem?id=3268

有一个点X开party 求其他所有点  去的路程和返程的和的最小值中的最大值

这里有一个小技巧 

总是把X当做原点 

返程是正常的最短路

去程就把路径翻转 求最短路 

这样只需要两次求最短路 

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <queue>
 5 #include <algorithm>
 6 #define MAXV 1007
 7 #define MAXE 100007
 8 #define INF 0x3f3f3f3f
 9 using namespace std;
10 
11 typedef pair<int,int> P ;//first 是距离 second是编号
12 struct Edge
13 {
14     int to, cost, next;
15     Edge() {}
16     Edge(int to, int cost, int next) : to(to), cost(cost), next(next) {}
17 }edge1[MAXE], edge2[MAXE];
18 
19 int head1[MAXV], head2[MAXV];
20 int num1 = 0, num2 = 0;
21 void Add1(int from, int to, int cost)
22 {
23     edge1[num1] = Edge(to, cost, head1[from]);
24     head1[from] = num1++;
25 }
26 
27 void Add2(int from, int to, int cost)
28 {
29     edge2[num2] = Edge(to, cost, head2[from]);
30     head2[from] = num2++;
31 }
32 
33 int N, M, X;
34 
35 int dijkstra(int s,int dist[], int head[], Edge edge[])
36 {
37     //int dist[MAXV];
38     priority_queue<P> que;
39     fill(dist, dist+MAXV, INF);
40     dist[s] = 0;
41     que.push(P(0, s));
42     while (!que.empty())
43     {
44         P p = que.top();
45         que.pop();
46         int t = head[p.second];
47         if (dist[p.second] < p.first) continue;
48         while (t != -1)
49         {
50             Edge e = edge[t];
51             if(dist[e.to] > dist[p.second] + e.cost)
52             {
53                 dist[e.to] = dist[p.second] + e.cost;
54                 que.push(P(dist[e.to], e.to));
55             }
56             t = e.next;
57         }
58     }
59 }
60 
61 int main()
62 {
63     int ans[MAXV];
64     int dist1[MAXV], dist2[MAXV];
65     freopen("in.txt", "r", stdin);
66     scanf("%d%d%d", &N, &M, &X);
67     memset(head1, -1, sizeof(head1));
68     memset(head2, -1, sizeof(head2));
69     memset(edge1, -1, sizeof(edge1));
70     memset(edge2, -1, sizeof(edge2));
71     memset(ans, 0, sizeof(ans));
72     for (int i = 0; i < M; i++)
73     {
74         int from, to, cost;
75         scanf("%d%d%d", &from, &to, &cost);
76         Add1(from, to, cost);
77         Add2(to, from, cost);
78     }
79     dijkstra(X, dist1, head1, edge1);//正常返回的路径
80     dijkstra(X, dist2, head2, edge2);//路径翻转之后 得到的dist就是 i到达X的路径
81     for (int i = 1; i <= N; i++)
82     {
83         ans[i] = dist1[i] + dist2[i];
84     }
85     sort(ans+1, ans+N+1);
86     printf("%d\n", ans[N]);
87     return 0;
88 }