BZOJ1097: [POI2007]旅游景点atr

用dijstra求出K个点两两的距离和1到他们，他们到N的距离，然后就可以状压了
可以预处理一下哪些点一定要在i之前走到，转成二进制存在pre[i]，这样可以加速判断

code：

#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long long#define inf INT_MAXusing namespace std;inline void read(int &x){    char c; while(!((c=getchar())>='0'&&c<='9'));    x=c-'0';    while((c=getchar())>='0'&&c<='9') (x*=10)+=c-'0';}inline void down(int &x,const int &y){if(x>y)x=y;}const int maxn = 41000;const int maxm = 210000;int n,m,K;bool v[25][25];int D[25][25];struct edge{    int y,c,nex;    edge(){}    edge(const int _y,const int _c,const int _nex){y=_y;c=_c;nex=_nex;}}a[maxm<<1]; int len,fir[maxn];inline void ins(const int x,const int y,const int c){a[++len]=edge(y,c,fir[x]);fir[x]=len;}struct node{    int x,i;    node(){}    node(const int _x,const int _i){x=_x;i=_i;}};inline bool operator <(const node &x,const node &y){return x.x>y.x;}priority_queue<node>q;int d[maxn];void Search(int s){    for(int i=1;i<=n;i++) d[i]=inf;    d[s]=0;    q.push(node(0,s));    while(!q.empty())    {        const node tmp=q.top(); q.pop();        if(tmp.x<d[tmp.i]) continue;        const int x=tmp.i;        for(int k=fir[x];k;k=a[k].nex)        {            const int y=a[k].y;            if(d[y]>d[x]+a[k].c)            {                d[y]=d[x]+a[k].c;                q.push(node(d[y],y));            }        }    }}int pre[25];int f[1<<21][21];int main(){    scanf("%d%d%d",&n,&m,&K);    for(int i=1;i<=m;i++)    {        int x,y,c; read(x); read(y); read(c);        ins(x,y,c); ins(y,x,c);    }    if(!K) { Search(1); printf("%d\n",d[n]); return 0; }    int ki; read(ki);    for(int i=1;i<=ki;i++)    {        int x,y; read(x); read(y);         v[x-1][y-1]=true;    }    for(int k=1;k<=K;k++)    {        for(int i=1;i<=K;i++) if(i!=k&&v[i][k])            for(int j=1;j<=K;j++) if(j!=k&&j!=i&&v[k][j])                v[i][j]=true;    }    for(int i=1;i<=K;i++)    {        for(int j=1;j<=K;j++) if(v[j][i]) pre[i]|=1<<j-1;    }    for(int i=2;i<=K+1;i++)    {        Search(i);        for(int j=1;j<=K;j++) D[i-1][j]=d[j+1];        D[i-1][K+1]=d[n];    }Search(1);    for(int i=0;i<(1<<K);i++) for(int j=1;j<=K;j++) f[i][j]=inf;    for(int i=1;i<=K;i++)        if(pre[i]==0) f[1<<i-1][i]=d[i+1];    for(int i=1;i<(1<<K);i++)    {        for(int j=1;j<=K;j++) if((i&(1<<j-1))&&f[i][j]==inf)        {            int tj=i^(1<<j-1);            if((tj&pre[j])==pre[j])            {                for(int k=1;k<=K;k++) if(j!=k&&f[tj][k]!=inf)                if(f[i][j]>f[tj][k]+D[k][j])                {                    f[i][j]=f[tj][k]+D[k][j];                    //g[i][j]=tj; gk[i][j]=k;                }            }        }    }    int ans=inf,al=(1<<K)-1;    for(int i=1;i<=K;i++)         if(f[al][i]!=inf) down(ans,f[al][i]+D[i][K+1]);    printf("%d\n",ans);    return 0;}