【PAT】【Advanced Level】1018. PBM (30)

来源：互联网发布：moment.js 计算时间差编辑：程序博客网时间：2024/06/01 08:35

原题链接：

https://www.patest.cn/contests/pat-a-practise/1018

思路：

DFS，同时得到结果的时候判断是否为最优

注意：

最优性的判断：距离>带出数目>带回数目

CODE：

#include<iostream>#include<cstring>#include<string>#define N 510using namespace std;int c,n,p,m;int dist[N][N];int cy[N];int ca=0;int path[N];int re[N];int mind=1000000;int minb=1000000;int mino=1000000;int floor=0;int ggg=0;bool fl[N];void dfs(int now,int dis){    //cout<<now<<" "<<car<<" "<<dis<<" "<<floor<<endl;    if (now==p)    {        int sum=0;        int mmm=0;        int sum1=0;        for (int i=1;i<=floor;i++)        {            sum1+=cy[path[i]];            sum+=c-cy[path[i]];            mmm=max(mmm,sum);        }        int mini=max(0,sum1+mmm-floor*c);        //cout<<now<<" "<<car<<" "<<dis<<" "<<floor<<endl;        if (dis<mind)        {            mind=dis;            for (int i=0;i<=floor;i++)                re[i]=path[i];            ggg=floor;            mino=mmm;            minb=mini;        }        else if (dis==mind)        {            if (mmm<mino)            {                mind=dis;                for (int i=0;i<=floor;i++)                    re[i]=path[i];                ggg=floor;                mino=mmm;                minb=mini;            }            else if(mmm==mino)            {                if (mini<minb)                {                    mind=dis;                    for (int i=0;i<=floor;i++)                        re[i]=path[i];                    ggg=floor;                    mino=mmm;                    minb=mini;                }            }        }        return;    }    for (int i=1;i<=n;i++)    {        if (fl[i]==1) continue;        if (dist[now][i]==-1) continue;        //cout<<now<<" "<<i<<" "<<car<<" "<<(cy[i]-c)<<endl;        floor++;        fl[i]=1;        path[floor]=i;        dfs(i,dis+dist[now][i]);        floor--;        fl[i]=0;    }    return;}int main(){    memset(dist,-1,sizeof(dist));    memset(fl,0,sizeof(fl));    cin>>c>>n>>p>>m;    c/=2;    for (int i=1;i<=n;i++)   cin>>cy[i];    //for (int i=1;i<=n;i++) cout<<cy[i]<<endl;    for (int i=0;i<m;i++)    {        int u,v,co;        cin>>u>>v>>co;        dist[u][v]=co;        dist[v][u]=co;    }    //cout<<endl;    fl[0]=1;    path[0]=0;    dfs(0,0);    cout<<mino<<" ";    for (int i=0;i<ggg;i++)        cout<<re[i]<<"->";    cout<<re[ggg]<<" ";    cout<<minb;    //cout<<max(sum1+mmm-c*ggg,0);    return 0;}

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