2018HDU多校补题 HDU 6386 (双端队列+虚点)

题意：

给定一张图以及每条路的类型，问从1到n至少需要换几次路可以走到。图的大小是$n=1e5$，$m=2e5$

题解：

首先想到这道题的最暴力的做法即，把每条边看出一个点，对于相邻的两条边连边，若这两条边类型相同则边权为$0$，否则边权为$1$。但是这样的化图的大小就是$m^2$级别的了，所以想到建立虚点，第一种虚点即结点，建立$<n,m+n,1>$,$<m+n,n,0>$的双向边，来连接相邻边，代价为1。第二种虚点为处于可一步到达的几个点，这种虚点只需要跑$dfs$对于枚举到的边向虚点连边权为0的边 即$<m+n,ant,0>$,$<ant,m+n,0>$,图建立完成后就是求最短路的过程，由于这张图只有边权$1$和$0$因此考虑双端队列求最短路即可。

代码：

/**
 *     author:     TelmaZzzz
 *     create:     2019-07-11-19.50.30
**/
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <ctime>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/priority_queue.hpp>
#include <deque>
using namespace __gnu_pbds;
#include <random>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
#define PB push_back
#define MP make_pair
#define INF 1073741824
#define inf 1152921504606846976
#define pi 3.14159265358979323846
//#pragma comment(linker,"/STACK:10240000,10240000")
mt19937 rand_(time(0));
const int N=8e5+7,M=4e6;
const long long mod=1e9+7;
inline int read(){int ret=0;char ch=getchar();bool f=1;for(;!isdigit(ch);ch=getchar()) f^=!(ch^'-');for(;isdigit(ch);ch=getchar()) ret=(ret<<1)+(ret<<3)+ch-48;return f?ret:-ret;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}//逆元
int wei[M];
int Head[N];
int head[N],NEXT[M],ver[M],tot;void link(int u,int v,int w){ver[++tot]=v;NEXT[tot]=head[u];head[u]=tot;wei[tot]=w;}
bool vis[N],Vis[N],mark[N];
int dis[N];
int ant,n;
void Link(int u,int v,int w){
    ver[++tot]=v;
    NEXT[tot]=Head[u];
    Head[u]=tot;
    wei[tot]=w;
}
void bfs(){
    for(int i=1;i<=ant;i++){
        dis[i]=INF;
        vis[i]=false;
        Vis[i]=false;
    }
    dis[1]=0;
    deque<int>q;
    q.push_front(1);
    while(q.size()){
        int x=q.front();
        //cout<<x<<endl;
        q.pop_front();
        if(vis[x]) continue;
        vis[x]=true;
        if(x==n) return;
        for(int i=Head[x];i;i=NEXT[i]){
            int y=ver[i];
            if(vis[y]) continue;
            dis[y]=min(dis[x]+wei[i],dis[y]);
            if(wei[i]==1){
                if(Vis[y]) continue;
                Vis[y]=true;
                q.push_back(y);
            }
            else {
                q.push_front(y);
                Vis[y]=true;
            }
        }
    }
}
void dfs(int x,int in){
    for(int i=head[x];i;i=NEXT[i]){
        if(mark[i/2]) continue;
        if(wei[i]!=in) continue;
        mark[i/2]=true;
        Link(ant,i/2+n,0);
        Link(i/2+n,ant,0);
        dfs(ver[i],in);
    }
}
int main(){
    //freopen("1.txt","r",stdin);
    //ios::sync_with_stdio(false);
    int m;
    int u,v,w;
    while(~scanf("%d%d",&n,&m)){
        memset(head,0,sizeof(head));
        memset(Head,0,sizeof(Head));
        memset(mark,false,sizeof(mark));
        tot=1;
        ant=n+m;
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&u,&v,&w);
            link(u,v,w);
            link(v,u,w);
        }
        for(int i=1;i<=m;i++){
            Link(ver[i*2],i+n,1);
            Link(ver[i*2+1],i+n,1);
            Link(i+n,ver[i*2],0);
            Link(i+n,ver[i*2+1],0);
            if(mark[i]) continue;
            ant++;
            dfs(ver[i*2],wei[i*2]);
        }
        bfs();
        if(dis[n]==INF){
            puts("-1");
        }
        else {
            printf("%d\n",dis[n]);
        }
    }
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}