题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4179
思路:不知道怎么回事,wa了n多次,然后不知道怎么回事就过了==,还是简单的说一下思路吧:一次以起点为源点跑一遍spfa,然后以终点为起点跑一次spfa,这样我们就可以枚举difficult为maxdist的边了,设该边的端点为x,y,于是有ans=min(ans,dist1[x]+Get_Dist(x,y)+dist2[y])。
1 #include
2 #include
3 #include
4 #include
5 #include
6 #include
7 #include
8 using namespace std;
9 typedef pair<int,double>Pair;
10 #define MAXN 44444
11 #define inf 1e16
12 double dist1[MAXN],dist2[MAXN];
13 bool mark[MAXN];
14 struct Edge{
15 int v;
16 double w;
17 Edge(){}
18 Edge(int _v,double _w){
19 v&#61;_v,w&#61;_w;
20 }
21 };
22 vector
23 vector<int>vet;
24 struct Point{
25 int x,y,z;
26 }point[MAXN];
27
28 struct Node{
29 int vs,vt;
30 }node[MAXN];
31 int n,m,st,ed,maxdist;
32
33 double Get_Dist(int vs,int vt){
34 double xx&#61;1.0*(point[vs].x-point[vt].x)*(point[vs].x-point[vt].x);
35 double yy&#61;1.0*(point[vs].y-point[vt].y)*(point[vs].y-point[vt].y);
36 double zz&#61;1.0*(point[vs].z-point[vt].z)*(point[vs].z-point[vt].z);
37 return sqrt(xx&#43;yy&#43;zz);
38 }
39
40 int Get(int vs,int vt){
41 if(point[vs].z<point[vt].z){
42 double xx&#61;1.0*(point[vs].x-point[vt].x)*(point[vs].x-point[vt].x);
43 double yy&#61;1.0*(point[vs].y-point[vt].y)*(point[vs].y-point[vt].y);
44 double dd&#61;sqrt(xx&#43;yy);
45 return (int)(100*(point[vt].z-point[vs].z)/dd);
46 }
47 return 0;
48 }
49
50 void SPFA(int st,vector
51 {
52 memset(mark,false,(n&#43;2)*sizeof(mark[0]));
53 for(int i&#61;1;i<&#61;n;i&#43;&#43;)dist[i]&#61;inf;
54 dist[st]&#61;0;mark[st]&#61;true;
55 queue<int>Q;
56 Q.push(st);
57 while(!Q.empty()){
58 int u&#61;Q.front();
59 Q.pop();
60 mark[u]&#61;false;
61 for(int i&#61;0;i