2015弱校联盟(1) - I. Travel

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;}