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;
}