BZOJ 1097: [POI2007]旅游景点atr 状态压缩+Dijkstra

我不会说我还没调过样例~

以后有时间的话再调吧~

Code: 

#include <bits/stdc++.h>  
#define N 20003 
#define ll long long  
#define inf 1000000000 
#define M (1<<21) 
#define setIO(s) freopen(s".in","r",stdin) 
using namespace std; 
ll d[30][N],f[M][22];      
int n,m,K,edges;      
int done[N],hd[N],to[N<<1],nex[N<<1],val[N<<1],Log[22],phase[22];              
int count(int c) 
{
    int l=0; 
    for(;c;c>>=1) if(c&1) ++l; 
    return l;      
}
void getmin(ll &a,ll b) 
{
    if(b<a) a=b;        
}
void addedge(int u,int v,int c) 
{
    nex[++edges]=hd[u],hd[u]=edges,to[edges]=v,val[edges]=c;    
}
struct Node
{
    int u; 
    ll dis;       
    Node(int u=0,ll dis=0):u(u),dis(dis){}   
    bool operator<(Node b)const
    {
        return b.dis<dis;     
    }
};
priority_queue<Node>q;                   
void Dijkstra(int s) 
{  
    int i,v,u;     
    for(i=0;i<=n;++i) d[s][i]=inf,done[i]=0; 
    for(d[s][s]=0,q.push(Node(s,0));!q.empty();) 
    {
        Node e=q.top(); q.pop();   
        u=e.u; 
        if(done[u]) continue;  
        done[u]=1;  
        for(i=hd[u];i;i=nex[i]) 
        {
            v=to[i]; 
            if(d[s][v]>d[s][u]+val[i]) 
            { 
                d[s][v]=d[s][u]+val[i]; 
                q.push(Node(v,d[s][v]));      
            }
        }   
    }    
}
int main() 
{  
    int i,j; 
    setIO("input"); 
    scanf("%d%d%d",&n,&m,&K);       
    for(i=0;i<=20;++i) Log[i]=(1<<i);             
    for(i=1;i<=m;++i) 
    {
        int a,b,c; 
        scanf("%d%d%d",&a,&b,&c),addedge(a,b,c),addedge(b,a,c); 
    } 
    for(i=1;i<=K+1;++i) Dijkstra(i);   
    if(!K) 
    {
        printf("%lld\n",d[1][n]); 
        return 0; 
    }      
    int Q; 
    scanf("%d",&Q); 
    for(i=1;i<=Q;++i) 
    {
        int a,b;            
        scanf("%d%d",&a,&b); 
        if(a!=1) phase[b]+=Log[a-2];        
    }
    memset(f,0x3f,sizeof(f));                  
    for(int sta=1;sta<Log[K];++sta) 
    {
        if(count(sta)==1) 
        {   
            for(j=2;j<=K+1;++j) 
            {
                if(Log[j-2]&sta) 
                {
                    f[sta][j-2]=d[1][j];  
                    printf("%d %lld\n",j,f[sta][j-2]);   
                    break;                                  
                }
            }
        }   
        for(j=2;j<=K+1;++j) 
        {
            if(Log[j-2]&sta)    
            {
                for(int cc=2;cc<=K+1;++cc) 
                {
                    if( !(Log[cc-2]&sta) && ((sta & phase[cc]) == phase[cc]) )      
                        getmin(f[sta|Log[cc-2]][cc-2],f[sta][j-2]+d[j][cc]);                
                }
            }
        }
    }  
    ll ans=10000000000;   
    for(i=2;i<=K+1;++i) 
    { 
        getmin(ans, f[Log[K]-1][i-2]+d[i][n]);               
    }
    printf("%lld\n",ans);      
    return 0;    
}