【PAT甲级】个人做题记录之：1018 Public Bike Management (30 分)_1018 public bike management 最后一个点tle-优快云博客

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PAT甲级圆满考完了，这次也考了一个相对不错的成绩，之后可能就很少刷PAT题库了，慢慢将之前刷题的过程补充记录上来吧

1018 Public Bike Management

题目描述

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.
在这里插入图片描述

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 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:

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.
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: C_max(≤100), always an even number, is the maximum capacity of each station; N (≤500), the total number of stations; S_p, 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 C_i (i=1,⋯,N) where each C_i is the current number of bikes at S_irespectively. Then M lines follow, each contains 3 numbers: S_i, S_j, and T_ij which describe the time T_ij taken to move betwen stations S_i and S_j. 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−>⋯−>S_p. Finally after another space, output the number of bikes that we must take back to PBMC after the condition of S_p 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:

3 0->2->3 0

思路：

本题考察的是递归的应用，即更为复杂的DFS，需要用到剪枝与回溯。

对于这类问题，建议使用全局变量存储一些关键性数据，如下：

int map[MAXN][MAXN];//存储地图信息
int C,N,M,D;
int bike[MAXN];//存储自行车信息
bool visit[MAXN];//储存是否拜访过
vector<int> pathtep,path;//临时存储路径,与最后储存路径
int mindis=INF;//储存最短路径长度
int need=0,back=0;

之后将数据输入，本人习惯写一个input函数，用于输入数据：

关键点：因为要求从起点开始运送自行车，将路过的每一个点的自行车补充至perfect，所以在最开始，记录每一个点缺少多少自行车（例如，C=10，某一点i自行车数量为3，则记录该点bike[i]=3-(10)/2=-2）这样做的目的是为了之后DFS的时候，可以直接对这个数进行处理

void input()
{
   
	scanf("%d %d %d %d",&C,&N,&D,&M);
	for(int i=1;i<=N;i++)
	{
   
		scanf("%d",&bike[i]);
		bike[i]=bike[i]-C/2;
	}
	for(int i=0;i<=N;i++)
		visit[i]=false;
	for(int i=0;i<=N;i++)
		for(int j=0;j<=N;j++)
			map[i][j]=INF;
	for(int i=0;i<M