状态压缩or 最短路 poj2686

来源：互联网发布：德州扑克数据分析软件编辑：程序博客网时间：2024/06/02 07:03

自己居然做出来了。用状态压缩的。。

但是用djk却是MLE 啊，然后改了之后是TLE。。。

这用dp是因为这个DAG

状态dp

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<string>#include<cstring>#include<iomanip>#include<iostream>#include<stack>#include<cmath>#include<map>#include<vector>#define ll long long#define inf 0x3f3f3f3f#define INF 1e9#define bug1 cout<<"bug1"<<endl;#define bug2 cout<<"bug2"<<endl;#define bug3 cout<<"bug3"<<endl;using namespace std;#define sf scanf#define pf printf#define mem(a,b) memset(a,b,sizeof(a));const int maxn=10;int n,m,p,a,b;double ma[15];double dp[1<<maxn][35];int e[35][35];void solve(){    for(int i=0;i<=(1<<n);++i){        for(int j=0;j<=m;++j){            dp[i][j]=INF;        }    }    a--;b--;    dp[(1<<n)-1][a]=0;    double ans=INF;    for(int i=(1<<n)-1;i>=0;i--){//        for(int j=0;j<m;++j){//10*10000*10;            if(dp[i][j]==inf)continue;            for(int k=0;k<n;++k){                if(!(i&(1<<k)))continue;                for(int v=0;v<m;++v){if(e[j][v]==inf)continue;                    int nxt=i&(~(1<<k));                    dp[nxt][v]=min(dp[nxt][v],dp[i][j]+e[j][v]/ma[k]);                    if(v==b)ans=min(ans,dp[nxt][v]);                }            }        }    }    if(ans>=INF)puts("Impossible");    else pf("%0.3lf\n",ans);}void djk(){    }int main(){    while(~sf("%d%d%d%d%d",&n,&m,&p,&a,&b)&&n){        for(int i=0;i<n;++i){            sf("%lf",&ma[i]);        }        mem(e,inf);        for(int i=1;i<=p;++i){            int u,v,c;            sf("%d%d%d",&u,&v,&c);            u--;v--;            e[u][v]=e[v][u]=min(e[v][u],c);        }        solve();    }}

用djk却是错的。。

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<string>#include<cstring>#include<iomanip>#include<iostream>#include<stack>#include<cmath>#include<map>#include<vector>#define ll long long#define inf 0x3f3f3f3f#define INF 1e9#define bug1 cout<<"bug1"<<endl;#define bug2 cout<<"bug2"<<endl;#define bug3 cout<<"bug3"<<endl;using namespace std;#define sf scanf#define pf printf#define mem(a,b) memset(a,b,sizeof(a));const int maxn=10;int n,m,p,a,b;double ma[15];double dp[1<<maxn][35];int e[35][35];struct Node{    int st;    int u;    double time;    bool friend operator <(Node a,Node b){        return a.time>b.time;    }};int done[1<<maxn][35];void djk(){    a--;b--;    mem(done,0);    for(int i=0;i<=(1<<n);++i){        for(int j=0;j<=m;++j){            dp[i][j]=INF;        }    }    priority_queue<Node>q;    Node ss;ss.u=a;ss.st=(1<<n)-1;    ss.time=0;    dp[ss.st][a]=0;    q.push(ss);    while(!q.empty()){        Node top=q.top();q.pop();        int u=top.u;int st=top.st;double time=top.time;        if(u==b){            pf("%.3lf\n",time);return;        }        if(done[st][u])continue;        done[st][u]=1;        for(int i=0;i<m;++i){            if(e[u][i]==inf)continue;            for(int j=0;j<n;++j){                if(!((1<<j)&st))continue;                int nxtst=st&(~(1<<j));                if(dp[st][u]==INF)continue;                dp[nxtst][i]=min(dp[nxtst][i],dp[st][u]+e[u][i]/ma[j]);                Node nv;                nv.u=i;nv.time=dp[nxtst][i];nv.st=nxtst;                q.push(nv);            }        }    }    pf("Impossible\n");}int main(){    while(~sf("%d%d%d%d%d",&n,&m,&p,&a,&b)&&n){        for(int i=0;i<n;++i){            sf("%lf",&ma[i]);        }        mem(e,inf);        for(int i=1;i<=p;++i){            int u,v,c;            sf("%d%d%d",&u,&v,&c);            u--;v--;            e[u][v]=e[v][u]=min(e[v][u],c);        }        djk();    }}

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