PAT - 甲级 - 1018. Public Bike Management (30)(Dijkstra+DFS)
2017-10-30 17:56
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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.
Figure 1
Figure 1 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 S is the current number of bikes
stored at S. Given that the maximum capacity of each station is 10. To solve the problem at S3, we have 2 different shortest paths:
1. PBMC -> S1 -> S3. In this case, 4 bikes must be sent from PBMC, because we can collect 1 bike from S1 and then take 5 bikes to S3,
so that both stations will be in perfect conditions.
2. PBMC -> S2 -> S3. This path requires the same time as path 1, but only 3 bikes sent from PBMC and hence is the one that will be chosen.
Input Specification:
Each input file contains one test case. For each case, the first line contains 4 numbers: Cmax (<= 100), always an even number, is the maximum capacity of each station; N (<= 500), the total number of stations; Sp,
the index of the problem station (the stations are numbered from 1 to N, and PBMC is represented by the vertex 0); and M, the number of roads. The second line contains N non-negative numbers Ci(i=1,...N) where each Ci is
the current number of bikes at Si respectively. Then M lines follow, each contains 3 numbers: Si, Sj, and Tij which describe
the time Tij taken to move betwen stations Si and Sj. All the numbers in a line are separated by a space.
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->S1->...->Sp. Finally after another
space, output the number of bikes that we must take back to PBMC after the condition of Sp is adjusted to perfect.
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:
题目要求:
1.求出a点到b点的最短路,并且输出路径
2.如果最短路有多条则求出需要从起点携带自行车最少的那一条
3.如果不需要从起点带,则求出需要带回起点最少自行车的那一条
解决方法:
1.求最短路和输出路径可以使用Dijkstra算法并且存储每个节点的父节点,逆序输出即可。
2.可以遍历最短路的所有情况,找出最少的情况即可。
3.同2.
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.
Figure 1
Figure 1 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 S is the current number of bikes
stored at S. Given that the maximum capacity of each station is 10. To solve the problem at S3, we have 2 different shortest paths:
1. PBMC -> S1 -> S3. In this case, 4 bikes must be sent from PBMC, because we can collect 1 bike from S1 and then take 5 bikes to S3,
so that both stations will be in perfect conditions.
2. PBMC -> S2 -> S3. This path requires the same time as path 1, but only 3 bikes sent from PBMC and hence is the one that will be chosen.
Input Specification:
Each input file contains one test case. For each case, the first line contains 4 numbers: Cmax (<= 100), always an even number, is the maximum capacity of each station; N (<= 500), the total number of stations; Sp,
the index of the problem station (the stations are numbered from 1 to N, and PBMC is represented by the vertex 0); and M, the number of roads. The second line contains N non-negative numbers Ci(i=1,...N) where each Ci is
the current number of bikes at Si respectively. Then M lines follow, each contains 3 numbers: Si, Sj, and Tij which describe
the time Tij taken to move betwen stations Si and Sj. All the numbers in a line are separated by a space.
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->S1->...->Sp. Finally after another
space, output the number of bikes that we must take back to PBMC after the condition of Sp is adjusted to perfect.
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:
10 3 3 5 6 7 0 0 1 1 0 2 1 0 3 3 1 3 1 2 3 1
Sample Output:
3 0->2->3 0
题目要求:
1.求出a点到b点的最短路,并且输出路径
2.如果最短路有多条则求出需要从起点携带自行车最少的那一条
3.如果不需要从起点带,则求出需要带回起点最少自行车的那一条
解决方法:
1.求最短路和输出路径可以使用Dijkstra算法并且存储每个节点的父节点,逆序输出即可。
2.可以遍历最短路的所有情况,找出最少的情况即可。
3.同2.
#include<iostream> #include<cstdio> #include<vector> #define INF 999999999 using namespace std; int a, b, c; int Cmax, N, Sp, M; int dis[505]; // 距离 int vis[505]; // 访问标记 int G[505][505]; // 边权 int weight[505]; // bikes vector<int> path; // 结果路径 vector<int> pre[505]; // 最短路径 vector<int> temp; // 过程路径 int minneed = INF; int minback = INF; void dfs(int v){ if(v == 0){ temp.push_back(v); int need = 0, back = 0; for(int i = temp.size()-1; i >= 0; i--){ int id = temp[i]; if(weight[id] > 0){ back += weight[id]; } if(weight[id] < 0){ if(back <= (0-weight[id])){ need += (0-weight[id]) - back; back = 0; }else{ back += weight[id]; } } } if(need < minneed){ minneed = need; minback = back; path = temp; } else if(need == minneed){ if(back < minback){ minback = back; path = temp; } } temp.pop_back(); return ; } temp.push_back(v); for(int i = 0; i < pre[v].size(); i++){ dfs(pre[v][i]); } temp.pop_back(); } int main(){ // freopen("input.txt","r",stdin); while(scanf("%d%d%d%d",&Cmax, &N, &Sp, &M) != EOF){ fill(G[0], G[0] + 505 * 505, INF); fill(dis, dis + 505, INF); // input for(int i = 1; i <= N; i++){ scanf("%d",&weight[i]); weight[i] = weight[i] - Cmax/2; vis[i] = 0; } dis[0] = 0; for(int i = 1; i <= M; i++){ scanf("%d%d%d",&a,&b,&c); G[a][b] = G[b][a] = c; } // Dijkstra for(int i = 0; i <= N; i++){ int x = -1, minn = INF; for(int j = 0; j <= N; j++){ if(!vis[j] && dis[j] < minn){ x = j; minn = dis[j]; } } if(x == -1) break; vis[x] = 1; // pre用于存储最短路径 for (int y = 0; y <= N; y++){ if(!vis[y] && G[x][y] != INF){ if(dis[x] + G[x][y] < dis[y]){ dis[y] = dis[x] + G[x][y]; pre[y].clear(); pre[y].push_back(x); } else if(dis[x] + G[x][y] == dis[y]){ pre[y].push_back(x); } } } } // 遍历所有最短路径,求出符合条件的路径 dfs(Sp); cout<<minneed<<" 0"; for(int i = path.size()-2; i >= 0; i--){ cout<<"->"<<path[i]; } cout<<" "<<minback<<endl; } return 0; }
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