PTA-1018——Public Bike Management

发布时间：2020-12-14 05:13:54 所属栏目：大数据来源：网络整理

导读：题目: There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it to any other stations in the city. The Public Bike Management

题目:

There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it to any other stations in the city.

The Public Bike Management Center (PBMC) keeps monitoring the real-time capacity of all the stations. A station is said to be in?perfect?condition if it is exactly half-full. If a station is full or empty,PBMC will collect or send bikes to adjust the condition of that station to perfect. And more,all the stations on the way will be adjusted as well.

When a problem station is reported,PBMC will always choose the shortest path to reach that station. If there are more than one shortest path,the one that requires the least number of bikes sent from PBMC will be chosen.

The above figure illustrates an example. The stations are represented by vertices and the roads correspond to the edges. The number on an edge is the time taken to reach one end station from another. The number written inside a vertex?

PBMC ->?
PBMC ->?

Input Specification:

Each input file contains one test case. For each case,the first line contains 4 numbers:?

Output Specification:

For each test case,print your results in one line. First output the number of bikes that PBMC must send. Then after one space,output the path in the format:? $0. Finally after another space,output the number of bikes that we must take back to PBMC after the condition of?$

Note that if such a path is not unique,output the one that requires minimum number of bikes that we must take back to PBMC. The judge‘s data guarantee that such a path is unique.

Sample Input:

Sample Output:

3 0->2->3 0

分析:

dijkstra找最短路径集+DFS回溯找出最优路径。参考博客：点击前往

代码:

  1 #include<iostream>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<vector>
  5 #define INF 0x6FFFFFF
  6 using namespace std;
  7 int cmax,n,p,m;
  8 vector<vector<int> > cost(501,vector<int>(501,INF));
  9 vector<int> bikeNum(501,0);
 10 vector<int> d(501,INF);
 11 vector<bool> used(501,false);
 12 vector<vector<int> >allPath(501);    //记录最短路径上每一个结点的上一个结点（最短路可能有多条，因此上一个结点可能有多个） 
 13 
 14 void dijkstra(){    //dijkstra算法寻找最短路 
 15     d[0]=0;
 16     while(true){
 17         int v=-1;
 18         for(int u=0;u<=n;u++){
 19             if(!used[u]&&(v==-1||d[u]<d[v])){
 20                 v=u;
 21             }
 22         }
 23         if(v==-1){
 24             break;
 25         }
 26         used[v]=true;
 27         for(int u=0;u<=n;u++){
 28             if(d[u]>d[v]+cost[v][u]){
 29                 d[u]=d[v]+cost[v][u];
 30                 allPath[u].clear();
 31                 allPath[u].push_back(v);
 32             }else if(d[u]==d[v]+cost[v][u]){
 33                 allPath[u].push_back(v);
 34             }
 35         }
 36     }
 37 }
 38 
 39 vector<int> possiblePath;
 40 vector<int> bestPath;
 41 int minCollect=INF;
 42 int minSend=INF;
 43 
 44 void findBest(int u){    //DFS回溯，寻找最优路径 
 45     possiblePath.push_back(u);
 46     if(u==0){    //回溯到起点 
 47         int send=0;
 48         int collect=0;
 49         for(int i=possiblePath.size()-1;i>=0;i--){
 50             int index=possiblePath[i];
 51             if(index==0){
 52                 continue;
 53             }
 54             if(bikeNum[index]>cmax/2){        //如果该站点超过半满，收集多余 
 55                 collect+=bikeNum[index]-cmax/2;
 56             }else if(bikeNum[index]<cmax/2){    //如果该站点少于半满 
 57                 collect-=(cmax/2-bikeNum[index]);    //分发缺少的数量 
 58                 if(collect<0){        //如果沿途收集的不够发，从起点支出并把收集的清空 
 59                     send-=collect;
 60                     collect=0;     
 61                 }
 62             }
 63         }
 64         if(send<minSend){
 65             minSend=send;
 66             minCollect=collect;
 67             bestPath=possiblePath;
 68         }else if(send==minSend&&collect<minCollect){
 69             minCollect=collect;
 70             bestPath=possiblePath;
 71         }
 72         return;
 73     }
 74     
 75     for(int i=0;i<allPath[u].size();i++){
 76         findBest(allPath[u][i]);
 77         possiblePath.pop_back();
 78     }
 79 }
 80 
 81 int main(){
 82     cin>>cmax>>n>>p>>m;
 83     for(int i=1;i<=n;i++){
 84         cin>>bikeNum[i];
 85     }
 86     for(int i=0;i<m;i++){
 87         int a,b,t;
 88         cin>>a>>b>>t;
 89         cost[a][b]=t;
 90         cost[b][a]=t;
 91     }
 92     dijkstra(); 
 93     findBest(p);
 94     cout<<minSend<<" 0";
 95     for(int i=bestPath.size()-2;i>=0;i--){
 96         cout<<"->"<<bestPath[i];
 97     } 
 98     cout<<" "<<minCollect;
 99     return 0;
100 }

（编辑：李大同）

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