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PAT1087. All Roads Lead to Rome

时间:2015-03-01 10:28:52      阅读:192      评论:0      收藏:0      [点我收藏+]

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Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers N (2<=N<=200), the number of cities, and K, the total number of routes between pairs of cities; followed by the name of the starting city. The next N-1 lines each gives the name of a city and an integer that represents the happiness one can gain from that city, except the starting city. Then K lines follow, each describes a route between two cities in the format "City1 City2 Cost". Here the name of a city is a string of 3 capital English letters, and the destination is always ROM which represents Rome.

Output Specification:

For each test case, we are supposed to find the route with the least cost. If such a route is not unique, the one with the maximum happiness will be recommended. If such a route is still not unique, then we output the one with the maximum average happiness -- it is guaranteed by the judge that such a solution exists and is unique.

Hence in the first line of output, you must print 4 numbers: the number of different routes with the least cost, the cost, the happiness, and the average happiness (take the integer part only) of the recommended route. Then in the next line, you are supposed to print the route in the format "City1->City2->...->ROM".

Sample Input:
6 7 HZH
ROM 100
PKN 40
GDN 55
PRS 95
BLN 80
ROM GDN 1
BLN ROM 1
HZH PKN 1
PRS ROM 2
BLN HZH 2
PKN GDN 1
HZH PRS 1
Sample Output:
3 3 195 97
HZH->PRS->ROM

思路:Dij+DFS  

最少代价------第一尺度

最多快乐------第二尺度

最大的平均距离------第三尺度

其中本程序第一尺度在Dij中完成,第二和第三尺度在DFS中完成。

技术分享
  1 #include <iostream>
  2 #include <cstdio>
  3 #include <map>
  4 #include <vector>
  5 #include <cstdio>
  6 #include <cstring>
  7 #include <string>
  8 #include <algorithm>
  9 using namespace std;
 10 #define MAX 210
 11 const int INF=0x3fffffff;
 12 int G[MAX][MAX];
 13 bool visited[MAX];
 14 int d[MAX];
 15 int weight[MAX];
 16 int N,K;
 17 string str;
 18 map<string,int>str2num;
 19 map<int,string>num2str;
 20 int cnt=0;
 21 int numpath=0;
 22 int start,des;
 23 vector<int>pre[MAX];
 24 vector<int>tempath,path;  //用来解决DFS的遍历问题
 25 int hapiness=-1;
 26 double avghap=-1;
 27 
 28 int GetId(string str)
 29 {
 30     if(str2num.find(str)!=str2num.end())
 31        return str2num[str];
 32     else
 33     {
 34         num2str[cnt]=str;
 35         str2num[str]=cnt;
 36         return cnt++;
 37     }
 38 }
 39 void Dij()
 40 {
 41     d[start]=0;
 42     //需要进行初始化pre数组???不过这是为什么????
 43      
 44     for(int i=0;i<N;i++)
 45     {
 46         int u=-1,min=INF;
 47         for(int j=0;j<N;j++)
 48         {
 49             if(d[j]<min&&!visited[j])
 50             {
 51                 min=d[j];
 52                 u=j;
 53             }
 54         }
 55         if(u==-1)
 56            return ;
 57         visited[u]=true;
 58         for(int v=0;v<N;v++)
 59         {
 60             if(!visited[v]&&G[u][v]!=INF)
 61             {
 62                 if(d[u]+G[u][v]<d[v])
 63                 {
 64                     pre[v].clear();
 65                     pre[v].push_back(u);
 66                     d[v]=G[u][v]+d[u]; 
 67                 }
 68                 else if(d[u]+G[u][v]==d[v])
 69                 {
 70                     pre[v].push_back(u);
 71                 }
 72             }
 73         }
 74     }
 75 }
 76 void DFS(int id)
 77 {
 78     if(id==start)
 79     {
 80         numpath++;
 81         double wei=0;
 82         tempath.push_back(id);
 83         for(int i=0;i<tempath.size();i++)
 84         {
 85             wei+=weight[tempath[i]];
 86         }
 87         if(wei>hapiness)
 88         {
 89             hapiness=wei;
 90             avghap=(wei*1.0)/(tempath.size()-1);
 91             path=tempath;
 92         }
 93         else if(wei==hapiness)
 94         {
 95             if((wei*1.0)/(tempath.size()-1)>avghap)
 96             {
 97                 avghap=(wei*1.0)/(tempath.size()-1);
 98                 path=tempath;
 99             }
100         }
101         tempath.pop_back();
102         return;
103     }
104     tempath.push_back(id);
105     for(int i=0;i<pre[id].size();i++)
106     {
107         DFS(pre[id][i]);
108     }
109     tempath.pop_back();
110 }
111 
112 int main(int argc, char *argv[])
113 {
114     cin>>N>>K>>str;
115     fill(G[0],G[0]+MAX*MAX,INF);
116     fill(d,d+MAX,INF);
117     memset(visited,false,sizeof(visited));
118     start=GetId(str);
119     des=GetId("ROM");
120     weight[start]=0;
121     for(int i=0;i<N-1;i++)
122     {
123         string tem;
124         int hapiness;
125         cin>>tem>>hapiness;
126         int id=GetId(tem);
127         weight[id]=hapiness;
128     }
129     for(int i=0;i<K;i++)
130     {
131         string str1,str2;
132         int length;
133         cin>>str1>>str2>>length;
134         int id1=GetId(str1);
135         int id2=GetId(str2);
136         G[id1][id2]=G[id2][id1]=length;
137     }
138     Dij();
139     DFS(des);
140     printf("%d %d %d %d\n",numpath,d[des],hapiness,(int)avghap);
141     
142     for(int i=path.size()-1;i>=0;i--)
143     {
144         string tem=num2str[path[i]];
145         cout<<tem;
146         if(i!=0)
147            cout<<"->";
148          else
149             cout<<endl;
150     }
151     return 0;
152 }
View Code

 

PAT1087. All Roads Lead to Rome

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原文地址:http://www.cnblogs.com/GoFly/p/4306645.html

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