SCU 4444 Travel(bfs)

最新推荐文章于 2019-01-22 18:59:44 发布

原创最新推荐文章于 2019-01-22 18:59:44 发布 · 295 阅读

0 ·

CC 4.0 BY-SA版权

竞赛专栏收录该内容

87 篇文章

订阅专栏

本文探讨了在一个由城镇组成的国家中寻找从城镇1到城镇n的最短旅行时间问题。该问题涉及高速公路和铁路两种交通方式，每种方式对应不同的旅行时间。文章提供了具体的解决思路及代码实现。

Travel

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≤105,0≤m≤5⋅105,1≤a,b≤109). 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

Sample Output

    2
    3

题解：
若1到n没有高铁，则判断1到n坐高铁和b的大小关系，
如果有高铁并且a>b则判断有没有一个点通过普通铁路连接
1和n两点，若有则判断a和2*b的关系
代码：

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#define  ll long long
using namespace std;
int n,m;
ll a,b;
vector<int>v[100005];
struct node
{
	int v;
	ll t;
};
queue<node>q;
bool bo[100005];
ll min1(ll a,ll b)
{
	if(a>b)return b;
	return a;
}
int main()
{
	while(~scanf("%d %d %lld %lld",&n,&m,&a,&b))
	{
		while(!q.empty()) q.pop();
		for(ll i=0;i<100005;i++) v[i].clear();
		memset(bo,0,sizeof(bo));
		bool bb=0;
		for(ll i=1;i<=m;i++)
		{
			int x,y;
			scanf("%d %d",&x,&y);
			v[x].push_back(y);
			v[y].push_back(x);
			if((x==1&&y==n)||(x==n&&y==1))bb=1;
		}
		if(a>=b)
		{
			if(bb==0)printf("%d\n",b);
			else
			{
				for(int i=2;i<n;i++)
				{
					bool is=0;
					for(int j=0;j<v[i].size();j++)
					{
						if(v[i][j]==1||v[i][j]==n)is=1;
					}
					if(is==0)
					{
						bb=0;break;
					}
				}
				if(bb==0)printf("%d\n",min(a,2*b));
				else printf("%d\n",a);
			}
			continue;
		}
		node s;
		s.v=1;
		s.t=0;
		q.push(s);
		ll mx=b;
		bo[1]=1;
		bool f=0;
		while(!q.empty())
		{
			node s;
			s=q.front();
			q.pop();
			for(int i=0;i<v[s.v].size();i++)
			{
				int y=v[s.v].at(i);
				if(bo[y]) continue;
				if(y==n)
				{
					mx=min1(mx,(s.t+1)*a);
					f=1;
					break;
				}
				node ss;
				ss.t=s.t+1;
				ss.v=y;
				q.push(ss);
			}
			if(f)break;
		}
		printf("%lld\n",mx);
	}
	return 0;
}