POJ 3411

用mark[i][st]记录到达i点时到所有点情况为st时的最小花费，向下一步走的时候可以查看要到的点需要经过的中间点是否在st中，然后用优先队列+最短路便可以解决了。

#include<cstdio>

#include<cstring>

#include<queue>

using namespace std;

const int N=12;

int head[N],nc;

struct edge

{

    int to,by,c1,c2,nxt;

}edge[N*3];

void add(int a,int b,int c,int c1,int c2)

{

    edge[nc].to=b;edge[nc].nxt=head[a];edge[nc].by=c;edge[nc].c1=c1;edge[nc].c2=c2;head[a]=nc++;

}

struct data

{

    int id,cost,state;

    bool operator<(const data &next)const

    {

        return cost>next.cost;

    }

    data(){}

    data(int _id,int _cost,int _state){id=_id;cost=_cost;state=_state;}

};

int mark[N][1<<N];

priority_queue<data> Q;

int dfs(int n)

{

    while(!Q.empty())

        Q.pop();

    memset(mark,-1,sizeof(mark));

    Q.push(data(1,0,1));

    int ans=-1;

    mark[1][1]=0;

    while(!Q.empty())

    {

        data a=Q.top();

        Q.pop();

        if(a.id==n)

        {

            if(ans==-1||ans>a.cost)

                ans=a.cost;

            continue;

        }

        for(int i=head[a.id];i!=-1;i=edge[i].nxt)

        {

            int t=edge[i].to;

            int st=a.state|(1<<(t-1));

            if((1<<(edge[i].by-1))&a.state)

            {

                if(mark[t][st]==-1||mark[t][st]>a.cost+edge[i].c1)

                {

                    mark[t][st]=a.cost+edge[i].c1;

                    Q.push(data(t,mark[t][st],st));

                }

            }

            if(mark[t][st]==-1||mark[t][st]>a.cost+edge[i].c2)

            {

                mark[t][st]=a.cost+edge[i].c2;

                Q.push(data(t,mark[t][st],st));

            }

        }

    }

    return ans;

}

int main()

{

    int n,m;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        memset(head,-1,sizeof(head));

        nc=0;

        for(int i=0;i<m;i++)

        {

            int a,b,c,c1,c2;

            scanf("%d%d%d%d%d",&a,&b,&c,&c1,&c2);

            add(a,b,c,c1,c2);

        }

        m=dfs(n);

        if(m==-1)

            printf("impossible\n");

        else

            printf("%d\n",m);

    }

    return 0;

}