PAT 1087. All Roads Lead to Rome (30) Dijkstra题——多条件最短，最短路数量，途径点

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题意:
给出一个图，除了给出每个节点间的距离外
还告知了每个节点的happy值
让你求一个最短路径
要求路程最短。
如果路程相同，则让路上累加的happy值最大
happy值相等，则让路上经过的城市最少。
并给出相同最小路程的路径数量。
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解题思路：
djkstra模板修改题
首先就是在dijksra中多加2种if判断，即判断h和c的
其次，求相同最小路径的数量。这个只要在更新时，若发现
  d[j] == d[select] + d[select][j]
  就给 roadnum[j] += roadnum[select]
  即这个点的途径路径数量，要增加，增加的就是你更新的这个select
还有就是求过程中经过的点，这个不需要用vector去存，弄一个pre[]前指向数组即可

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笔记：

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#include<iostream>
#include<stdio.h>
#include<string>
#include<vector>
#include<queue>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
#include<stack>
#include<map>
#include<set>
#include<unordered_map>
using namespace std;
#define M 300
#define INF 0x7ffffff

string cityname[M];
int happy[M];
int dis[M][M];
map<string,int > getid;


void dijk(int s, int e, int n){
  int i, j, dmin, hmax, select, cmin;
  int d[M], vis[M], h[M], rnum[M],c[M],pre[M];
  for(i = 0; i < n; i++){
    d[i] = INF;
    vis[i] = 0;
    h[i] = -1;
    rnum[i] = 0;
    c[i] = INF;
  }
  d[s] = 0;
  h[s] = 0;
  rnum[s] = 1;
  c[s] = 1;
  pre[s] = -1;
  for(i = 0;i < n; i++){
    dmin = INF;
    hmax = -1;
    cmin = INF;
    for(j = 0;j < n;j++){
      if(vis[j])
        continue;
      if(d[j] < dmin){
        dmin = d[j];
        hmax = h[j];
        cmin = c[j];
        select = j;
      }else if(d[j] == dmin){
        if(h[j] > hmax){
          hmax = h[j];
          cmin = c[j];
          select = j;
        }else if(h[j] == hmax){
          if(c[j] <  cmin){
            cmin = c[j];
            select = j;
          }
        }
      }
    }
    if(dmin ==INF)
      break;
    vis[select] = 1;
    for(j = 0;j < n; j++){
      if(vis[j])
        continue;
      if(d[select] + dis[select][j] < d[j]){
        d[j] = d[select] + dis[select][j];
        h[j] = h[select] + happy[j];
        c[j] = c[select] + 1;
        pre[j] = select;
        rnum[j] = rnum[select];
      }else if(d[select] + dis[select][j] == d[j]){
        rnum[j] += rnum[select]; //统计different least road number
        if(h[select] + happy[j] > h[j]){
          h[j] = h[select] + happy[j];
          c[j] = c[select] +1;
          pre[j] = select;
        }else if(h[select] + happy[j] == h[j]){
          if(c[select] + 1 <  c[j]){
            c[j] = c[select] + 1;
            pre[j] = select;
          }
        }
      }
    }
  }

  printf("%d %d %d %d\n",rnum[e], d[e], h[e], h[e]/(c[e] - 1));
  int p = e;
  vector<string> ans;
  while(p != -1){
    ans.push_back(cityname[p]);
    p = pre[p];
  }
  for(i = ans.size() - 1;i >= 0;i--){
    if(i != 0)
      cout<<ans[i]<<"->";
    else cout<<ans[i]<<endl;
    
  }
  
}


int main(){
  int i, n, m, d, j, end;
  string start, s, e;
  cin>>n>>m>>start;
  for(i=0;i<n;i++)
    for(j=0;j<n;j++)
      if(i!=j)
        dis[i][j] = INF;
      else dis[i][j] = 0;
  
  happy[0] = 0;
  cityname[0] = start;
  getid[start] = 0;
  
  for(i = 1; i < n ; i++){
    cin>>cityname[i]>>happy[i];
    if(cityname[i] == "ROM")
      end = i;
    getid[cityname[i]] = i;
  }
  while(m--){
    cin>>s>>e>>d;
    dis[getid[s]][getid[e]] = d;
    dis[getid[e]][getid[s]] = d;  
  }
  if(start == "ROM"){
    printf("1 0 0 0\n");
    cout<<"ROM"<<endl;
  }
  else
    dijk(0, end, n);
  return 0;
}