水题。。可是做的人很少。。。

<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">思路： 把每个点i拆成两个点i和i+n，所有与i相关的出边与i相连，与i相关的入边与i+n相连，在求s到t的最短路时就是求s到t+n的最短路，这样就能解决环的问题了。然后看一个二维的dis[i][j]，代表从s到i困难度最高为j的最短路，最后上Dijkstra，1A~</span>

<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);"></span><pre name="code" class="cpp">#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<cstdlib>

using namespace std;
const int maxn = 50005;
const int maxd = 25;
const double inf = 1e12;

struct Node{
    int a,b,dis;
    Node(int a,int b,int dis): a(a),b(b),dis(dis) {}
    bool operator < (const Node &z) const {
        return dis > z.dis;
    }
};

struct Point{
    double x,y,z;
}p[maxn];

struct Edge{
    int to,d;
    double cost;
};

int n,m,s,t,lv;
double dis[maxn][maxd];
vector<Edge> G[maxn];
priority_queue<Node> que;

void init(){
    while(!que.empty()) que.pop();
    for(int i=0;i<n;i++) G[i].clear();
}

int cal_d(int a,int b){
    if(p[a].z < p[b].z){
        double rise = p[b].z - p[a].z;
        double run = sqrt((p[a].x - p[b].x)*(p[a].x - p[b].x) + (p[a].y - p[b].y)*(p[a].y - p[b].y));
        return (int)(100*rise/run);
    }else return 0;
}

double cal_cost(int a,int b){
    return sqrt((p[a].x - p[b].x)*(p[a].x - p[b].x) + (p[a].y - p[b].y)*(p[a].y - p[b].y) + (p[a].z - p[b].z)*(p[a].z - p[b].z));
}

void add_Edge(int from,int to){
    Edge e;
    e.to = to + n;e.d = cal_d(from,to);e.cost = cal_cost(from,to);
    G[from].push_back(e);
    e.to = from + n;e.d = cal_d(to,from);;
    G[to].push_back(e);
}

void dij(){
    for(int i=0;i<2*n;i++){
        for(int j=0;j<=lv;j++){
            dis[i][j] = inf;
        }
    }
    dis[s][0] = 0;
    que.push(Node(s,0,0));

    while(!que.empty()){
        Node nd = que.top();que.pop();
        int v = nd.a,d = nd.b;
        if(dis[v][d] < nd.dis) continue;
        int u = v % n;
        for(int i=0;i<G[u].size();i++){
            Edge e = G[u][i];
            int k = max(d,e.d);
            if(k <= lv && dis[e.to][k] > dis[v][d] + e.cost){
                dis[e.to][k] = dis[v][d] + e.cost;
                que.push(Node(e.to,k,dis[e.to][k]));
            }
        }
    }
    if(dis[t+n][lv] == inf) printf("None\n");
    else printf("%.1f\n",dis[t+n][lv]);
}

int main(){
    while(~scanf("%d%d",&n,&m)){
        if(n + m == 0) break;
        init();
        for(int i=0;i<n;i++){
            scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
        }
        int a,b;
        while(m--){
            scanf("%d%d",&a,&b);
            a--;b--;
            add_Edge(a,b);
        }
        scanf("%d%d%d",&s,&t,&lv);
        s--;t--;
        dij();
    }
    return 0;
}