PAT 1018 Public Bike Management (Dijkstra + dfs记录多条最短路径)

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <vector>
  4 
  5 #define INF 1000000000
  6 
  7 using namespace std;
  8 
  9 int dist[510];
 10 vector<int> pre[510];
 11 vector<int> path, tmp;
 12 int g[510][510]; 
 13 int s[510];
 14 int vis[510];
 15 int C, N, Sp, M;
 16 int minNeed = INF;        //需要额外带的自行车
 17 int back = INF;          //需要带回去的自行车 
 18 
 19 /*
 20 10 3 3 3
 21 5 0 10
 22 0 1 1
 23 1 2 1
 24 2 3 1
 25 */
 26 
 27 //先假设最短路唯一，看看能得多少分
 28 void dijkstra (int start)
 29 {
 30     for(int i = 0; i <= N; ++i)
 31     {
 32         dist[i] = INF;
 33         vis[i] = 0;
 34     }
 35     dist[start] = 0;
 36     
 37     while(1)
 38     {
 39         int mark = -1, minDist = INF;
 40         for(int i = 0; i <= N; ++i)
 41         {
 42             if(!vis[i] && dist[i] < minDist)
 43             {
 44                 minDist = dist[i];
 45                 mark = i;
 46             }
 47         }
 48         
 49         if(mark == -1)
 50             break;
 51             
 52         vis[mark] = 1;
 53         
 54         for(int i = 0; i <= N; ++i)
 55         {
 56             if(!vis[i])
 57             {
 58                 if(dist[mark] + g[mark][i] < dist[i])
 59                 {
 60                     dist[i] = dist[mark] + g[mark][i];
 61                     pre[i].clear();
 62                     pre[i].push_back(mark);    // 记录路径 
 63                 }
 64                 else if(dist[mark] + g[mark][i] == dist[i])
 65                 {
 66                     pre[i].push_back(mark);    // 记录路径
 67                 }    
 68             }
 69         }
 70     }
 71     
 72 }
 73 
 74 void dfs(int v)
 75 {
 76     // 深度优先路过的顶点,存入tmp 
 77     tmp.push_back(v);
 78     // 当到达顶点0的时候，开始计算
 79     if(v == 0)
 80     {
 81         int perfectnum = C/2;
 82         int difference;
 83         int need = 0;        //需要带的自行车 
 84         int available = 0;    //可用的(多余的)自行车    
 85         for(int i = tmp.size()-2; i >= 0; --i)
 86         {
 87             if(s[tmp[i]] > perfectnum)     
 88             {
 89                 // 有单车多余 
 90                 difference = s[tmp[i]] - perfectnum;
 91                 available += difference;
 92             }
 93             else if(s[tmp[i]] < perfectnum)
 94             {
 95                 // 单车不够 
 96                 difference = perfectnum - s[tmp[i]];
 97                 if(available >= 0)
 98                 {
 99                     if(available > difference)
100                     {
101                         // 此时need数量不变 
102                         available -= difference;
103                     }
104                     else
105                     {
106                         need += difference - available; 
107                         available = 0;
108                     }
109                     
110                 }
111             }
112         }
113         
114         if(need < minNeed)
115         {
116             minNeed = need;
117             back = available;
118             path = tmp;
119         } else if(need == minNeed && available < back)
120         {
121             back = available;
122             path = tmp;
123         }
124         
125         // 顶点在dfs访问结束后弹出 
126         tmp.pop_back(); 
127         return;    
128     }
129     
130     for(int i = 0; i < pre[v].size(); ++i)
131         dfs(pre[v][i]);
132     
133     tmp.pop_back();        // 该顶点在dfs访问结束后弹出 
134 }
135 
136 int main()
137 {
138     scanf("%d%d%d%d", &C, &N, &Sp, &M);
139     for(int i = 1; i <= N; ++i)
140         scanf("%d", &s[i]);
141     
142     for(int i = 0; i <= N; ++i)
143     {
144         for(int j = 0; j <= N; ++j)
145         {
146             if(i == j)
147                 g[i][j] = 0;
148             else
149                 g[i][j] = INF;
150         }
151     }
152     
153     int a, b, cost;
154     for(int i = 0; i < M; ++i)
155     {
156         scanf("%d%d%d", &a, &b, &cost);
157         g[a][b] = g[b][a] = cost;
158     }
159     
160     dijkstra(0);
161     dfs(Sp);
162     
163     printf("%d 0", minNeed);
164     
165     int psize = path.size();
166     for(int i = psize-2; i >= 0; --i)
167     {
168         printf("->%d", path[i]);
169     }
170     
171     printf(" %d", back);
172 
173     return 0;
174 }