1087. All Roads Lead to Rome (30)

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//dijstra#include<cstdio>#include<algorithm>#include<string>#include<map>#include<vector>#include<iostream>using namespace std;#define maxn 1000#define INF  1000000000map<string, int> cityToNum;map<int, string> NumTocity;int n,m,G[maxn][maxn];int vis[maxn]={false},pre[maxn]={0};int num[maxn]={0},d[maxn],weight[maxn]={0},w[maxn],pt[maxn]={0};string st;void init(){  fill(G[0],G[0]+maxn*maxn,INF);  fill(d, d+maxn, INF);  fill(w,w+maxn, INF);}void Dijstra(int s){  d[s] = 0;  num[s] = 1;  w[s] = 0;    //n次循环  for(int i=0; i<n; i++)  {    int u=-1,MIN = INF;    for(int j=0; j<n; j++)    {      if(vis[j]==false && d[j]<MIN)      {        u = j;        MIN = d[j];      }    }    if(u == -1)      return;    vis[u] = true;    for(int v=0; v<n; v++)    {      if(vis[v]==false && G[u][v]!=INF)      {        //找到了话费最小的，全部更新        if(d[u]+G[u][v]<d[v])        {          d[v] = d[u]+G[u][v];          w[v] = w[u]+weight[v];          num[v] = num[u];          pre[v] = u;          pt[v] = pt[u] + 1;        }        else if(d[u]+G[u][v]==d[v])        {          num[v] += num[u];          //如果幸福值最多，选择更新          if(w[u]+weight[v]>w[v])          {            w[v] = w[u]+weight[v];            pre[v] = u;            pt[v] = pt[u]+1;          }          //如果等于，计算平均点权          else if(w[u]+weight[v]==w[v])          {            double uavg = (w[u]+weight[v])*1.0/(pt[u]+1);            double vavg = (w[v])*1.0/pt[v];            if(uavg>vavg)            {              pt[v] = pt[u]+1;              pre[v] = u;            }          }        }      }    }  }}void showPath(int s, int v){  while (s == v)  {    cout<<NumTocity[s];    return;  }  showPath(s,pre[v]);  cout<<"->"<<NumTocity[v];}int main(){  string city1,city2;  //输入数据  cin>>n>>m>>st;  init();  cityToNum[st] = 0;  NumTocity[0] = st;  //构造图  for(int i=1; i<n; i++)  {    cin>>city1>>weight[i];    cityToNum[city1] = i;    NumTocity[i] = city1;  }  for(int i=0; i<m; i++)  {    int x;    cin>>city1>>city2>>x;    int a = cityToNum[city1];    int b = cityToNum[city2];    G[a][b] = x;    G[b][a] = x;  }  //遍历  Dijstra(0);  //输出结果  int ed = cityToNum["ROM"];  printf("%d %d %d %d\n",num[ed],d[ed],w[ed],w[ed]/pt[ed]);  showPath(0,ed);  return 0;}
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