hdu 4179(有限制的最短路）

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<iostream>

  2 #include<cstdio>

  3 #include<cstring>

  4 #include<algorithm>

  5 #include<queue>

  6 #include<vector>

  7 #include<cmath>

  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=_v,w=_w;

 20     }

 21 };

 22 vector<Edge>map1[MAXN],map2[MAXN];

 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=1.0*(point[vs].x-point[vt].x)*(point[vs].x-point[vt].x);

 35     double yy=1.0*(point[vs].y-point[vt].y)*(point[vs].y-point[vt].y);

 36     double zz=1.0*(point[vs].z-point[vt].z)*(point[vs].z-point[vt].z);

 37     return sqrt(xx+yy+zz);

 38 }

 39 

 40 int Get(int vs,int vt){

 41     if(point[vs].z<point[vt].z){

 42         double xx=1.0*(point[vs].x-point[vt].x)*(point[vs].x-point[vt].x);

 43         double yy=1.0*(point[vs].y-point[vt].y)*(point[vs].y-point[vt].y);

 44         double dd=sqrt(xx+yy);

 45         return (int)(100*(point[vt].z-point[vs].z)/dd);

 46     }

 47     return 0;

 48 }

 49 

 50 void SPFA(int st,vector<Edge>map[],double dist[])

 51 {

 52     memset(mark,false,(n+2)*sizeof(mark[0]));

 53     for(int i=1;i<=n;i++)dist[i]=inf;

 54     dist[st]=0;mark[st]=true;

 55     queue<int>Q;

 56     Q.push(st);

 57     while(!Q.empty()){

 58         int u=Q.front();

 59         Q.pop();

 60         mark[u]=false;

 61         for(int i=0;i<map[u].size();i++){

 62             int v=map[u][i].v;

 63             double w=map[u][i].w;

 64             if(dist[u]+w<dist[v]){

 65                 dist[v]=dist[u]+w;

 66                 if(!mark[v]){ mark[v]=true;Q.push(v); }

 67             }

 68         }

 69     }

 70 }

 71 

 72 

 73 int main()

 74 {

 75    // freopen("1.txt","r",stdin);

 76     while(scanf("%d%d",&n,&m),(n+m)){

 77         for(int i=1;i<=n;i++){ map1[i].clear();map2[i].clear();};

 78         vet.clear();

 79         for(int i=1;i<=n;i++){

 80             scanf("%d%d%d",&point[i].x,&point[i].y,&point[i].z);

 81         }

 82         for(int i=1;i<=m;i++){

 83             scanf("%d%d",&node[i].vs,&node[i].vt);

 84         }

 85         scanf("%d%d%d",&st,&ed,&maxdist);

 86         for(int i=1;i<=m;i++){

 87             double dd=Get_Dist(node[i].vs,node[i].vt);

 88             int d1=Get(node[i].vs,node[i].vt);

 89             int d2=Get(node[i].vt,node[i].vs);

 90             if(d1<=maxdist){

 91                 map1[node[i].vs].push_back(Edge(node[i].vt,dd));

 92                 map2[node[i].vt].push_back(Edge(node[i].vs,dd));

 93             }

 94             if(d2<=maxdist){

 95                 map1[node[i].vt].push_back(Edge(node[i].vs,dd));

 96                 map2[node[i].vs].push_back(Edge(node[i].vt,dd));

 97             }

 98             if(d1==maxdist){

 99                 vet.push_back(node[i].vs);

100                 vet.push_back(node[i].vt);

101             }

102             if(d2==maxdist){

103                 vet.push_back(node[i].vt);

104                 vet.push_back(node[i].vs);

105             }

106         }

107         SPFA(st,map1,dist1);

108         SPFA(ed,map2,dist2);

109         double ans=inf;

110         for(int i=0;i<vet.size();i+=2){

111             int x=vet[i],y=vet[i+1];

112             double dd=dist1[x]+Get_Dist(x,y)+dist2[y];

113             if(dd<ans)ans=dd;

114         }

115         if(ans<inf){

116             printf("%.1lf\n",ans);

117         }else

118             puts("None");

119     }

120     return 0;

121 }

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