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最新推荐文章于 2016-08-22 14:30:51 发布

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最新推荐文章于 2016-08-22 14:30:51 发布

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原文链接：http://www.cnblogs.com/lj030/p/3200906.html

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Dijkstra最短路

模拟每个key的坠落时间，发现就是Dijkstra。

求出每个key的时间求其最大值，再求每条边整个坠落的时间，求其最大值，得二者最大值即结果。

其中每条边的坠落事件为tv+(cost-(tv-tu))/2,其中u，v为边的两个key节点，tv>tu，cost为边的权值

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <queue>
 5 using namespace std;
 6 const int maxn=500+10;
 7 const int inf=0x3f3f3f3f;
 8 const int maxm=maxn*maxn;
 9 struct Edge
10 {
11     int to,next,cost,from,dis;
12 };
13 struct HeapNode
14 {
15     int d,u;
16     bool operator < (const HeapNode& rhs) const { return d>rhs.d; }
17 };
18 Edge edge[maxm];
19 int head[maxn],dis[maxn],vis[maxn];
20 int n,m,tot,loc;
21 double maxv;
22 void dijkstra()
23 {
24     priority_queue<HeapNode> q;
25     memset(dis,inf,sizeof(dis));
26     memset(vis,0,sizeof(vis));
27     dis[1]=0;
28     maxv=0;
29     loc=1;
30     q.push((HeapNode){0,1});
31     while(!q.empty())
32     {
33         HeapNode x=q.top(); q.pop();
34         int u=x.u;
35         if(vis[u]) continue;
36         if(dis[u]>maxv)
37         {
38             maxv=dis[u];
39             loc=u;
40         }
41         vis[u]=1;
42         for(int i=head[u];i!=-1;i=edge[i].next)
43         {
44             if(dis[edge[i].to]>(dis[u]+edge[i].cost))
45             {
46                 dis[edge[i].to]=dis[u]+edge[i].cost;
47                 q.push((HeapNode){dis[edge[i].to],edge[i].to});
48             }
49         }
50     }
51 }
52 int main()
53 {
54     int cas=0;
55     while(scanf("%d%d",&n,&m))
56     {
57         if(!n&&!m) break;
58         tot=0;
59         memset(head,-1,sizeof(head));
60         int i;
61         int u,v,cost;
62         for(i=0;i<m;i++)
63         {
64             scanf("%d%d%d",&u,&v,&cost);
65             edge[tot].cost=cost; edge[tot].to=v; edge[tot].from=u;
66             edge[tot].next=head[u]; head[u]=tot++;
67             edge[tot].cost=cost; edge[tot].to=u; edge[tot].from=v;
68             edge[tot].next=head[v]; head[v]=tot++;
69         }
70         dijkstra();
71         int loc1,loc2,flag=0;
72         double tem;
73         for(i=0;i<tot;i+=2)
74         {
75             tem=(edge[i].cost+dis[edge[i].from]+dis[edge[i].to])/2.0;
76             if(tem>maxv)
77             {
78                 flag=1;
79                 maxv=tem;
80                 loc1=min(edge[i].from,edge[i].to);
81                 loc2=max(edge[i].from,edge[i].to);
82             }
83         }
84         printf("System #%d\n",++cas);
85         if(flag) printf("The last domino falls after %.1f seconds, between key dominoes %d and %d.\n",maxv,loc1,loc2);
86         else printf("The last domino falls after %.1f seconds, at key domino %d.\n",maxv,loc);
87         printf("\n");
88     }
89     return 0;
90 }

转载于:https://www.cnblogs.com/lj030/p/3200906.html