1087 All Roads Lead to Rome (30 分)难度: 一般 / Dijkstra

Posted 2021-12-12 辉小歌

https://pintia.cn/problem-sets/994805342720868352/problems/994805379664297984
PAT这种类型地题出了很多次了。
思路:

• 先将字符串映射到数字，数字映射到字符串。便于两者直接的查找。
• 跑一下Dijkstra
• dfs一下，走一下所有的最短路，维护题目所要求的信息。

#include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
int g[N][N],w[N],dist[N],vis[N],n,k;
int idx,st,ed;
int cnt,max_happy,avg_happy;
string start;
unordered_map<string,int>mp;
unordered_map<int,string>hush;
unordered_map<int,int>path,ans;
vector<int>ve;
int get(string x)
{
    if(mp.count(x)==0) mp[x]=++idx;
    hush[mp[x]]=x;
    return mp[x];
}
void Dijkstra(int st)
{
    memset(dist,0x3f,sizeof dist);
    dist[st]=0;
    for(int i=0;i<n;i++)
    {
        int t=-1;
        for(int j=1;j<=n;j++) if(!vis[j]&&(t==-1 || dist[j]<dist[t])) t=j;
        vis[t]=1;
        for(int j=1;j<=n;j++) dist[j]=min(dist[j],dist[t]+g[t][j]);
    }
}
void dfs(int u,int fa,int sum,int len)
{
    if(u==ed)
    {
        cnt++;
        if(sum>max_happy)
        {
            max_happy=sum;
            avg_happy=sum/len;
            ans=path;
        }else if(sum==max_happy&&sum/len>avg_happy) 
        {
            avg_happy=sum/len,ans=path;
        }
        return;
    }
    for(int i=1;i<=n;i++)
    {
        if(i==fa) continue;//避免往回走
        if(dist[i]==dist[u]+g[u][i]) //说明是最短路上的点
        {
            path[i]=u;
            dfs(i,u,sum+w[i],len+1);
        }
    }
}
int main(void)
{
    memset(g,0x3f,sizeof g);
    cin>>n>>k>>start;
    for(int i=1;i<=n-1;i++)
    {
        string name;
        int c; cin>>name>>c;
        int u=get(name);
        w[u]=c;
    }
    for(int i=1;i<=k;i++)
    {
        string x,y;
        int a,b,c; cin>>x>>y>>c;
        a=get(x),b=get(y);
        g[a][b]=g[b][a]=min(g[a][b],c);
    }
    st=get(start),ed=get("ROM");
    Dijkstra(st);
    dfs(st,-1,0,0);
    cout<<cnt<<" "<<dist[ed]<<" "<<max_happy<<" "<<avg_happy<<endl;
    do
    {
        ve.push_back(ed);
        ed=ans[ed];
    }while(ed!=st);
    cout<<start;
    for(int i=ve.size()-1;i>=0;i--) cout<<"->"<<hush[ve[i]];
    return 0;
}