BZOJ2622 深入虎穴（最短路径）

　　如果对某个点能求出与其相邻的所有点到达出口的最短时间，那么该点的答案就可以在其中取次小值了。

　　对于dijkstra魔改一下就能做到这个。初始时将所有出口的最短时间设为0并放入堆，记录最短和次短路径，每个点更新后将次短路径入堆。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 100010
#define M 2000010
#define inf 700000000
int n,m,k,p[N],d[N],d2[N],t=0,ans=inf;
bool flag[N];
struct data{int to,nxt,len;
}edge[M];
struct data2
{
    int x,d;
    bool operator <(const data2&a) const
    {
        return d>a.d;
    }
};
priority_queue<data2> q;
void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;}
void dijkstra()
{
    memset(flag,0,sizeof(flag));
    for (int i=1;i<=n;i++)
    {
        while (!q.empty()&&flag[q.top().x]) q.pop();
        if (q.empty()) break;
        data2 v=q.top();q.pop();
        flag[v.x]=1;
        for (int j=p[v.x];j;j=edge[j].nxt)
        if (v.d+edge[j].len<d[edge[j].to])
        {
            d2[edge[j].to]=d[edge[j].to];
            d[edge[j].to]=v.d+edge[j].len;
            q.push((data2){edge[j].to,d2[edge[j].to]});
        }
        else if (v.d+edge[j].len<d2[edge[j].to])
        {
            d2[edge[j].to]=v.d+edge[j].len;
            q.push((data2){edge[j].to,d2[edge[j].to]});
        }
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj2622.in","r",stdin);
    freopen("bzoj2622.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    n=read(),m=read(),k=read();
    for (int i=1;i<=m;i++)
    {
        int x=read()+1,y=read()+1,z=read();
        addedge(x,y,z),addedge(y,x,z);
    }
    memset(d,42,sizeof(d));memset(d2,42,sizeof(d2));
    for (int i=1;i<=k;i++)
    {
        int x=read()+1;
        d[x]=d2[x]=0;q.push((data2){x,0});
    }
    dijkstra();
    cout<<d2[1];
    return 0;
}