I. Travel

Time Limit: 3000ms

Memory Limit: 65536KB

The country frog lives in has n towns which are conveniently numbered by 1,2,…,n.

Among n(n−1)/2 pairs of towns, m of them are connected by bidirectional highway, which needs a minutes to travel. The other pairs are connected by railway, which needs b minutes to travel.

Find the minimum time to travel from town 1 to town n.

Input

The input consists of multiple tests. For each test:

The first line contains 4 integers n,m,a,b (2≤n≤10^5,0≤m≤5*10^5,1≤a,b≤10^9). Each of the following m lines contains 2 integers ui,vi, which denotes cities ui and vi are connected by highway. (1≤ui,vi≤n,ui≠vi).

Output

For each test, write 1 integer which denotes the minimum time.

Sample Input

3 2 1 3

1 2

2 3

3 2 2 3

1 2

2 3

Sample Output

2

3

对于(1,n);

(1)如果之间是铁路,则需要判断公路是不是更快

(2)如果是公路,则需要判断铁路是不是更快

分别bfs一次;

#include <bits/stdc++.h>
#define LL long long
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout) using namespace std; const int INF = 0x3f3f3f3f; const int Max = 1e5+100; typedef struct node
{
int x;
int num;
} Node; int n,m; LL A,B; LL Dist[Max]; vector<int>Pn[Max]; bool vis[Max];
bool visb[Max];
void init()
{
for(int i=1; i<=n; i++)
{
Pn[i].clear();
vis[i]=false;
} } void bfsa()//公路
{
queue<int>Q;
int b;
vis[1]=true;
Dist[n]=INF;
Dist[1]=0;
Q.push(1);
while(!Q.empty())
{
b=Q.front();
Q.pop();
int ans = Pn[b].size();
for(int i=0; i<ans; i++)
{
if(!vis[Pn[b][i]])
{ if(Dist[b]+A<=B)
{
Dist[Pn[b][i]]=Dist[b]+A;
vis[Pn[b][i]]=true;
Q.push(Pn[b][i]);
}
else
{
return ;
}
if(Pn[b][i]==n)
{
return ;
}
}
}
}
}
void bfsb()//铁路
{
queue<int>Q;
int b;
Dist[n]=INF;
Dist[1]=0;
vis[1]=true;
Q.push(1);
while(!Q.empty())
{
b=Q.front();
Q.pop();
for(int i=1; i<=n; i++)
{
visb[i]=false;
}
int ans = Pn[b].size();
for(int i=0; i<ans; i++)
{
visb[Pn[b][i]]=true;
}
for(int i=1; i<=n; i++)
{
if(!visb[i]&&!vis[i])
{
if(Dist[b]+B<=A)
{
vis[i]=true;
Dist[i]=Dist[b]+B;
Q.push(i);
}
else
{
return ;
}
if(i==n)
{
return ;
}
}
}
}
} int main()
{
int u,v;
int Dis;
while(~scanf("%d %d %lld %lld",&n,&m,&A,&B))
{
init();
Dis=-1;
for(int i=1; i<=m; i++)
{
scanf("%d %d",&u,&v);
if((u==1&&v==n)||(u==n&&v==1))
{
Dis=A;
}
Pn[u].push_back(v);
Pn[v].push_back(u);
}
if(Dis==-1)
{
bfsa();
printf("%lld\n",min(B,Dist[n]));
}
else
{
bfsb();
printf("%lld\n",min(A,Dist[n]));
}
}
return 0;
}

2015弱校联盟(1) - I. Travel的更多相关文章

  1. 2015弱校联盟(2) - J. Usoperanto

    J. Usoperanto Time Limit: 8000ms Memory Limit: 256000KB Usoperanto is an artificial spoken language ...

  2. 2015弱校联盟(1) - C. Censor

    C. Censor Time Limit: 2000ms Memory Limit: 65536KB frog is now a editor to censor so-called sensitiv ...

  3. 2015弱校联盟(1) - B. Carries

    B. Carries Time Limit: 1000ms Memory Limit: 65536KB frog has n integers a1,a2,-,an, and she wants to ...

  4. 2015弱校联盟(1) -J. Right turn

    J. Right turn Time Limit: 1000ms Memory Limit: 65536KB frog is trapped in a maze. The maze is infini ...

  5. 2015弱校联盟(1) -A. Easy Math

    A. Easy Math Time Limit: 2000ms Memory Limit: 65536KB Given n integers a1,a2,-,an, check if the sum ...

  6. 2015弱校联盟(1) - E. Rectangle

    E. Rectangle Time Limit: 1000ms Memory Limit: 65536KB 64-bit integer IO format: %lld Java class name ...

  7. 2016弱校联盟十一专场10.2---Around the World(深搜+组合数、逆元)

    题目链接 https://acm.bnu.edu.cn/v3/problem_show.php?pid=52305 problem  description In ICPCCamp, there ar ...

  8. (2016弱校联盟十一专场10.3) D Parentheses

    题目链接 把左括号看成A右括号看成B,推一下就行了.好久之前写的,推到最后发现是一个有规律的序列. #include <bits/stdc++.h> using namespace std ...

  9. (2016弱校联盟十一专场10.3) B.Help the Princess!

    题目链接 宽搜一下就行. #include <iostream> #include<cstdio> #include<cstring> #include<qu ...

随机推荐

  1. ArcGIS AddIN开发之自定义鼠标样式

    如果想修改Windows默认的鼠标样式,可以这样 //设置鼠标样式 this.Cursor = System.Windows.Forms.Cursors.Cross; 可是如果想设置成一些自定义的很好 ...

  2. MVC概念性的内容

    MVC:    是一个缩写(model + view + control),      Model:是一些类文件,  功能:负责增删改查, 负责跟数据库打交道 (把数据存入到数据库: 从数据库把数据读 ...

  3. 【iCore3 双核心板】例程八:定时器PWM实验——呼吸灯

    实验指导书及代码包下载: http://pan.baidu.com/s/1dEnH5dB iCore3 购买链接: https://item.taobao.com/item.htm?id=524229 ...

  4. ios-获取通讯录 姓名和电话

    #import "ViewController.h" #import <ContactsUI/ContactsUI.h> @interface ViewControll ...

  5. json.parse 与 json.stringfy

    转自 :http://blog.csdn.net/wangxiaohu__/article/details/7254598 parse用于从一个字符串中解析出json对象,如 var str = '{ ...

  6. 获取 Cookie

    /// <summary>    /// 获取WPF url 地址中的Cookies    /// </summary>    public partial class Coo ...

  7. jQuery/Javascript 事件停止冒泡

    Demo: <div id='wrap'> <button id='btn'>btn</button> </div> 一般情况下,两个元素分别添加点击事 ...

  8. Wordpress制作文章页面single.php

    可以调用的文章内容: 调用文章标题:<?php the_title(); ?> 调用文章内容:<?php the_content(); ?> 调用文章摘要:<?php t ...

  9. 多线程 - CyclicBarrier

    一个同步辅助类,它允许一组线程互相等待,直到到达某个公共屏障点 (common barrier point).在涉及一组固定大小的线程的程序中,这些线程必须不时地互相等待,此时 CyclicBarri ...

  10. Oracle中用户的基本操作

    创建用户 1.首先登陆到系统用户sys(sys用户具有创建用户的权限). 2.然后在代码编辑框写入创建用户的代码. 语法:CREATE USER user_name IDENTIFIED BY pas ...