PAT_A1087#All Roads Lead to Rome

Source:

PAT A1087 All Roads Lead to Rome (30 分)

Description:

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), 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 Klines 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 recommanded. 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 recommanded 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

Keys:

• 最短路径

Attention:

• 结点编号从1开始，因为映射判定的条件是结点编号==0

Code:

  1 /*
  2 Data: 2019-06-18 15:47:57
  3 Problem: PAT_A1087#All Roads Lead to Rome
  4 AC: 37:38
  5 
  6 题目大意：
  7 从起点至终点ROM花费最少且最幸福的一条路，
  8 若不唯一，给出平均幸福指数最高的路径
  9 
 10 输入：
 11 第一行给出，城市数N，路径数K，起点城市
 12 接下来N-1行，城市名，幸福指数
 13 接下来K行，v1,v2,cost
 14 输出：
 15 最短路径数目，花费，幸福指数，平均幸福指数（整数部分,除去起点）
 16 起点至终点的路径
 17 */
 18 #include<cstdio>
 19 #include<vector>
 20 #include<string>
 21 #include<iostream>
 22 #include<map>
 23 #include<algorithm>
 24 using namespace std;
 25 const int M=210,INF=1e9;
 26 int grap[M][M],vis[M],d[M],w[M];
 27 int n,s,t,pt=1,cnt=0,optValue=0,optAvg=0;
 28 map<string,int> mp;
 29 string city[M],ROM="ROM";
 30 vector<int> pre[M],temp,opt;
 31 
 32 int Str(string s)
 33 {
 34     if(mp[s]==0)
 35     {
 36         mp[s]=pt;
 37         return pt++;
 38     }
 39     else
 40         return mp[s];
 41 }
 42 
 43 void Dijskra(int st)
 44 {
 45     fill(d,d+M,INF);
 46     fill(vis,vis+M,0);
 47     d[st]=0;
 48     for(int i=1; i<=n; i++)
 49     {
 50         int u=-1,Min=INF;
 51         for(int j=1; j<=n; j++)
 52         {
 53             if(vis[j]==0 && d[j]<Min)
 54             {
 55                 u = j;
 56                 Min = d[j];
 57             }
 58         }
 59         if(u==-1)   return;
 60         vis[u]=1;
 61         for(int v=1; v<=n; v++)
 62         {
 63             if(vis[v]==0 && grap[u][v]!=INF)
 64             {
 65                 if(d[u]+grap[u][v] < d[v])
 66                 {
 67                     d[v] = d[u] + grap[u][v];
 68                     pre[v].clear();
 69                     pre[v].push_back(u);
 70                 }
 71                 else if(d[u]+grap[u][v] == d[v])
 72                     pre[v].push_back(u);
 73             }
 74         }
 75     }
 76 }
 77 
 78 void DFS(int v)
 79 {
 80     if(v == s)
 81     {
 82         cnt++;
 83         int value=0;
 84         for(int i=temp.size()-1; i>=0; i--)
 85             value += w[temp[i]];
 86         if(value > optValue)
 87         {
 88             optValue = value;
 89             optAvg = value/temp.size();
 90             opt = temp;
 91         }
 92         else if(value == optValue)
 93         {
 94             if(value/temp.size() > optAvg)
 95             {
 96                 optAvg = value/temp.size();
 97                 opt = temp;
 98             }
 99         }
100         return;
101     }
102 
103     temp.push_back(v);
104     for(int i=0; i<pre[v].size(); i++)
105         DFS(pre[v][i]);
106     temp.pop_back();
107 }
108 
109 int main()
110 {
111 #ifdef ONLINE_JUDGE
112 #else
113     freopen("Test.txt", "r", stdin);
114 #endif // ONLINE_JUDGE
115 
116     fill(grap[0],grap[0]+M*M,INF);
117     fill(w,w+M,0);
118 
119     string name;
120     int v1,v2,m;
121     scanf("%d%d", &n,&m);
122     cin >> name;
123     s = Str(name);
124     city[s]=name;
125     for(int i=1; i<n; i++)
126     {
127         cin >> name;
128         v1 = Str(name);
129         city[v1]=name;
130         if(name==ROM)
131             t = v1;
132         scanf("%d", &w[v1]);
133     }
134     for(int i=0; i<m; i++)
135     {
136         cin >> name;
137         v1 = Str(name);
138         cin >> name;
139         v2 = Str(name);
140         scanf("%d", &grap[v1][v2]);
141         grap[v2][v1]=grap[v1][v2];
142     }
143     Dijskra(s);
144     DFS(t);
145     printf("%d %d %d %d\n", cnt,d[t],optValue,optAvg);
146     cout << city[s];
147     for(int i=opt.size()-1; i>=0; i--)
148         cout << "->" << city[opt[i]];
149 
150     return 0;
151 }

 

转载于:https://www.cnblogs.com/blue-lin/p/11045608.html