SCU 4444: Travel(最短路)

Travel

The country frog lives in has n

towns which are conveniently numbered by 1,2,,n

.

Among n(n1)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 (2n105,0m5105,1a,b109). Each of the following m lines contains 2 integers ui,vi, which denotes cities ui and vi are connected by highway. (1ui,vin,uivi

).

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


分两种情况:
1. 1到 n 铁路可以直连 跑一次最短路取 Min(b,dist[n])
2. 1 到n 高速公路连通 那么我们BFS跑最短路 ,每次构一次新图,每个点入队一次.因为是完全图所以每次可以入队的点都非常多.所以可以暴力.

#include <iostream>
#include <cstdio>
#include <string.h>
#include <queue>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long LL;
const LL INF = 1e16;
const LL N = 100005;
struct Edge{
    LL v,next;
    LL w;
}edge[10*N];
LL head[N];
LL tot,n,m,a,b;
bool vis[N];
LL low[N];
LL MIN ;
void addEdge(LL u,LL v,LL w,LL &k){
    edge[k].v = v,edge[k].w = w,edge[k].next = head[u],head[u]=k++;
}
struct Node{
    int u;
    int step;
};
void init(){
    memset(head,-1,sizeof(head));
    tot = 0;
}
void spfa(LL pos){
    for(LL i=1;i<=n;i++){
        low[i] = INF;
        vis[i] = false;
    }
    low[pos] = 0;
    queue<LL>q;
    while(!q.empty()) q.pop();
    q.push(pos);
    while(!q.empty()){
        LL u = q.front();
        q.pop();
        vis[u] = false;
        for(LL k=head[u];k!=-1;k=edge[k].next){
            LL w = edge[k].w,v = edge[k].v;
            if(low[v]>low[u]+w){
                low[v] = low[u]+w;
                if(!vis[v]){
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
}
bool vis1[N];
int bfs(int pos){
    memset(vis,0,sizeof(vis));
    queue<Node> q;
    Node node;
    vis[pos] = 1;
    node.u = pos;
    node.step = 0;
    q.push(node);
    while(!q.empty()){
        memset(vis1,false,sizeof(vis1));
        node = q.front();
        int u = node.u;
        int step = node.step;
        /// if((step+1)*b>a) return -1;  TLE
        q.pop();
        vis1[u] = 1;
        for(int k=head[u];k!=-1;k=edge[k].next){
            int v = edge[k].v;
            vis1[v] = true;
        }
        Node next;
        if(!vis1[n]) return step+1;
        for(int i=n;i>=1;i--){
            if(!vis1[i]&&!vis[i]){
                if((step+1)*b>a) return -1;
                next.u = i;
                next.step = step+1;
                q.push(next);
                vis[i] = true;
            }
        }
    }
    return -1;
}
int main(){
    while(scanf("%lld%lld%lld%lld",&n,&m,&a,&b)!=EOF){
        init();
        bool flag = false;
        for(int i=0;i<m;i++){
            LL u,v;
            scanf("%lld%lld",&u,&v);
            addEdge(u,v,a,tot);
            addEdge(v,u,a,tot);
            if(u==1&&v==n||u==n&&v==1) flag = 1;
        }
        LL ans = 0;
        if(!flag){
            spfa(1);
            ans = min(low[n],b);
        }else{
            int step = bfs(1);
            if(step==-1) ans = a;
            else ans = min(a,step*b);
        }
        printf("%lld\n",ans);
    }
}

 

 
posted @ 2017-08-22 10:41  樱花庄的龙之介大人  阅读(...)  评论(... 编辑 收藏