POJ3268

题意：给你n个奶牛，m个农场，在x农场开派对，问你所有奶牛中必须花费最大的时间

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 #include <queue>
 7 using namespace std;
 8 #define M 100005
 9 #define N 1005
10 #define inf 9999999
11 int size1,size2,n,m,x;
12 int dis[N],vis[N],head1[N],head2[N];
13 int sum[N];
14 struct Edge
15 {
16     int v,w,next;
17     Edge(){}
18     Edge(int V,int W,int NEXT):v(V),w(W),next(NEXT){}
19 }edge1[M],edge2[M];
20 
21 void Init()
22 {
23     size1 = 0;
24     size2 = 0;
25     memset(head1,-1,sizeof(head1));
26     memset(head2,-1,sizeof(head2));
27 }
28 
29 void InsertEdge1(int u,int v,int w)
30 {
31     edge1[size1] = Edge(v,w,head1[u]);
32     head1[u] = size1++;
33 }
34 void InsertEdge2(int u,int v,int w)
35 {
36     edge2[size2] = Edge(v,w,head2[u]);
37     head2[u] = size2++;
38 }
39 
40 void spfa(int st,Edge edge[],int head[])
41 {
42     for(int i=1; i<=n; i++)
43     {
44         vis[i] = 0;
45         dis[i] = inf;
46     }
47     dis[st] = 0; vis[st] = 1;
48     queue<int> Q;
49     Q.push(st);
50     while(!Q.empty())
51     {
52         int u = Q.front();
53         Q.pop();
54         vis[u] = 0;
55         for(int i=head[u]; i!=-1; i = edge[i].next)
56         {
57             int v = edge[i].v;
58             if(dis[v] > dis[u] + edge[i].w)
59             {
60                 dis[v] = dis[u] + edge[i].w ;
61                 if(!vis[v])
62                 {
63                     vis[v] = 1;
64                     Q.push(v);
65                 }
66             }
67         }
68     }
69 }
70 
71 int main()
72 {
73     while(scanf("%d%d%d",&n,&m,&x)!=EOF)
74     {
75         Init();
76         int u,v,w;
77         for(int i=0; i<m; i++)
78         {
79             scanf("%d%d%d",&u,&v,&w);
80             InsertEdge1(u,v,w);
81             InsertEdge2(v,u,w);
82         }
83         memset(sum,0,sizeof(sum));
84         spfa(x,edge1,head1);
85         for(int i=1; i<=n; i++)
86         sum[i] += dis[i];
87         spfa(x,edge2,head2);
88         for(int i=1; i<=n; i++)
89         sum[i] += dis[i];
90         int ans = -1;
91         for(int i=1; i<=n; i++)
92         ans = max(ans,sum[i]);
93         printf("%d\n",ans);
94     } 
95     return 0;
96 }

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转载于:https://www.cnblogs.com/ar940507/p/3244909.html