• 131072K
 

One day in the jail, F·F invites Jolyne Kujo (JOJO in brief) to play tennis with her. However, Pucci the father somehow knows it and wants to stop her. There are NN spots in the jail and MM roads connecting some of the spots. JOJO finds that Pucci knows the route of the former (K-1)(K−1)-th shortest path. If Pucci spots JOJO in one of these K-1K−1 routes, Pucci will use his stand Whitesnake and put the disk into JOJO's body, which means JOJO won't be able to make it to the destination. So, JOJO needs to take the KK-th quickest path to get to the destination. What's more, JOJO only has TT units of time, so she needs to hurry.

JOJO starts from spot SS, and the destination is numbered EE. It is possible that JOJO's path contains any spot more than one time. Please tell JOJO whether she can make arrive at the destination using no more than TT units of time.

Input

There are at most 5050 test cases.

The first line contains two integers NN and MM (1 \leq N \leq 1000, 0 \leq M \leq 10000)(1≤N≤1000,0≤M≤10000). Stations are numbered from 11 to NN.

The second line contains four numbers S, E, KS,E,K and TT ( 1 \leq S,E \leq N1≤S,E≤N, S \neq ES≠E, 1 \leq K \leq 100001≤K≤10000, 1 \leq T \leq 1000000001≤T≤100000000 ).

Then MM lines follows, each line containing three numbers U, VU,V and WW (1 \leq U,V \leq N, 1 \leq W \leq 1000)(1≤U,V≤N,1≤W≤1000) . It shows that there is a directed road from UU-th spot to VV-th spot with time WW.

It is guaranteed that for any two spots there will be only one directed road from spot AA to spot BB (1 \leq A,B \leq N, A \neq B)(1≤A,B≤N,A≠B), but it is possible that both directed road <A,B><A,B> and directed road <B,A><B,A>exist.

All the test cases are generated randomly.

Output

One line containing a sentence. If it is possible for JOJO to arrive at the destination in time, output "yareyaredawa" (without quote), else output "Whitesnake!" (without quote).

样例输入复制

2 2
1 2 2 14
1 2 5
2 1 4

样例输出复制

yareyaredawa

题目来源

ACM-ICPC 2018 沈阳赛区网络预赛

//图上两点之间的第k最短路径的长度

 #define  P pair<int,int>
#define ph push_back
#define N 1009
#define M 10009
const int inf =0x3f3f3f3f;
int n,m,s,e,k,t;
int dis[N];
vector<P>ve[N],se[N];
void init()
{
for(int i=;i<n;i++) {
ve[i].clear();
se[i].clear();
}
}
struct Node{
int to,w;
bool operator <(const Node&a)const{
return w+dis[to]>a.w+dis[a.to];
}
};
void dijk()
{
priority_queue<P,vector<P>,greater<P> >que;
for(int i=;i<N;i++) dis[i]=inf;
dis[e]=;
que.push({,e});
while(!que.empty()){
P temp=que.top();que.pop();
int v=temp.second;
if(dis[v]<temp.first) continue;
for(int i=;i<se[v].size();i++){//反向边
P p=se[v][i];
int x=p.first,w=p.second;
if(dis[x]>dis[v]+w){
dis[x]=dis[v]+w;
que.push({dis[x],x});
}
}
}
}
int astar()
{
priority_queue<Node>que;
que.push({s,});
k--;
while(!que.empty()){
Node temp=que.top();que.pop();
int v=temp.to;
if(v==e) {
if(k) k--;
else return temp.w;
}
for(int i=;i<ve[v].size();i++){
P p=ve[v][i];
int x=p.first,w=p.second;
que.push({x,w+temp.w});
}
}
return -;
}
int main()
{
while(~scanf("%d%d",&n,&m)){
scanf("%d%d%d%d",&s,&e,&k,&t);
int u,v,w;
for(int i=;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
ve[u].ph({v,w});
se[v].ph({u,w});
}
dijk();
if(dis[s]==inf) printf("Whitesnake!\n");
else{
if(s==e) k++;//k+1
int ans=astar();
if(ans<=t&&ans!=-) printf("yareyaredawa\n");
else{
printf("Whitesnake!\n");
}
}
}
return ;
}

图上两点之间的第k最短路径的长度 ACM-ICPC 2018 沈阳赛区网络预赛 D. Made In Heaven的更多相关文章

  1. ACM-ICPC 2018 沈阳赛区网络预赛 D Made In Heaven(第k短路,A*算法)

    https://nanti.jisuanke.com/t/31445 题意 能否在t时间内把第k短路走完. 分析 A*算法板子. #include <iostream> #include ...

  2. ACM-ICPC 2018 沈阳赛区网络预赛 D. Made In Heaven(第k短路模板)

    求第k短路模板 先逆向求每个点到终点的距离,再用dij算法,不会超时(虽然还没搞明白为啥... #include<iostream> #include<cstdio> #inc ...

  3. ACM-ICPC 2018 沈阳赛区网络预赛 D. Made In Heaven(约束第K短路)

    题意:求11到nn的第kk短的路径长度,如果超过TT输出Whitesnake!Whitesnake!,否则输出yareyaredawayareyaredawa. 好无以为 , 这就是一道模板题, 当是 ...

  4. ACM-ICPC 2018 沈阳赛区网络预赛 K Supreme Number(规律)

    https://nanti.jisuanke.com/t/31452 题意 给出一个n (2 ≤ N ≤ 10100 ),找到最接近且小于n的一个数,这个数需要满足每位上的数字构成的集合的每个非空子集 ...

  5. ACM-ICPC 2018 沈阳赛区网络预赛-D:Made In Heaven(K短路+A*模板)

    Made In Heaven One day in the jail, F·F invites Jolyne Kujo (JOJO in brief) to play tennis with her. ...

  6. ACM-ICPC 2018 沈阳赛区网络预赛 F Fantastic Graph(贪心或有源汇上下界网络流)

    https://nanti.jisuanke.com/t/31447 题意 一个二分图,左边N个点,右边M个点,中间K条边,问你是否可以删掉边使得所有点的度数在[L,R]之间 分析 最大流不太会.. ...

  7. ACM-ICPC 2018 沈阳赛区网络预赛 F. Fantastic Graph (贪心或有源汇上下界网络流)

    "Oh, There is a bipartite graph.""Make it Fantastic."X wants to check whether a ...

  8. ACM-ICPC 2018 沈阳赛区网络预赛 F. Fantastic Graph(有源上下界最大流 模板)

    关于有源上下界最大流: https://blog.csdn.net/regina8023/article/details/45815023 #include<cstdio> #includ ...

  9. ACM-ICPC 2018 沈阳赛区网络预赛 Made In Heaven(K短路)题解

    思路:K短路裸题 代码: #include<queue> #include<cstring> #include<set> #include<map> # ...

随机推荐

  1. Codeforces Round #377 (Div. 2) D. Exams 贪心 + 简单模拟

    http://codeforces.com/contest/732/problem/D 这题我发现很多人用二分答案,但是是不用的. 我们统计一个数值all表示要准备考试的所有日子和.+m(这些时间用来 ...

  2. PS高级特训班 百度云资源(价值2180元)

    课程目录:   第1章第一期1第一节 火焰拳头1:12:252第二节 荷叶合成00:05:143第三节 新年巨惠海报(一)1:00:374第四节 新年巨惠海报(二)1:05:345第五节 美食印刷品1 ...

  3. Vsftp设置为PASV mode(被动模式传送)

    首先配置vsftpd.conf文件: #vi /etc/vsftpd/vsftpd.conf 在文件的末尾加上: pasv_enable=YES pasv_max_port=30010 pasv_mi ...

  4. WPF 模拟Button按钮事件触发

    this.Submit.AddHandler(Button.ClickEvent, new RoutedEventHandler(this.Submit_Click)); //这种是无效的方法 thi ...

  5. Unity3d网格合并

    几个不同的物体,在Unity3d中可以将网格合并在一起,用于优化. 在Unity3d中构建临时场景如下:建一C#脚本名为"CombineMeshes",挂在Cube上. Combi ...

  6. ParallelsDesktop安装DOS7.1并与MAC共享文件

    ParallelsDesktop安装DOS7.1并与MAC共享文件 Table of Contents 1. 在Parallels Desktop中安装DOS7.1 2. 配置与Mac共享文件 1 在 ...

  7. GoDaddy网站程序根目录 网站文件上传到虚拟主机哪个目录

    用的linux虚拟主机,网站根目录为public_html,(window主机的目录为httpdocs)我们需要把本地做好的网站上传到此目录下 cPanel控制面板 - 文件管理器 - public_ ...

  8. Android自定义组件系列【17】——教你如何高仿微信录音Toast

    一.Toast介绍 平时我们在Android开发中会经常用到一个叫Toast的东西,官方解释如下 A toast is a view containing a quick little message ...

  9. ubuntu下安装ffmpeg扩展

    可通过PPA进行安装 sudo add-apt-repository ppa:kirillshkrogalev/ffmpeg-next sudo apt-get update sudo apt-get ...

  10. DA层(数据访问层)的方法不用静态的

    1.静态方法,不会经过构造函数,所以你不能通过构造函数来初始参数,你只能通过传递参数,来初始他当你有多种参数需要传递的时候,你就要不断重载他了.当然你可以用参数型的类型,不过如果参数有一定结构,就很麻 ...