PTA甲级考试真题练习18——1018 Public Bike Management

题目

生词生字

1. The Public Bike Management Center (PBMC) keeps monitoring the real-time capacity of all the stations
   公共自行车管理中心（PBMC）持续监控所有车站的实时通行能力
2. A station is said to be in perfect condition if it is exactly half-full.
   一个站点如果正好处于半满状态被认为是完美的情况
3. The stations are represented by vertices and the roads correspond to the edges
   站点用顶点表示并且边表示路长

思路

较难的求图的单源最短路径题，难点在于题意，题意读懂就好办
题目主要是说从PBMC开始向目的地一步一步的走，如果当前站点差自行车，则立即补上，并且累加总的需求和，如果不差或者多余，把多余的自行车累加到下一个站点，直到最终站点，最终站点如果有多的则全部返回PBMC中，后面的站点如果多余是不能往前面补充的，最终按照最短路径<最小需求<最小返回进行排序，输出第一个
由题意可知，该题最佳算法是深度遍历不是Dijkstra算法，因为在深度遍历的过程中可以将每个站点的状态判断清楚，累加需求和并且将多余的传到下一个站点

#include <iostream>
#include<string.h>
#include<algorithm>
#include<vector>
#include<stack>
#include<map>
#include<set>
#include<queue>
#include<list>
using namespace std;
#define nmax 1000  //一个用作示例的小顺序表的最大长度
#define mint 999999

int graph[nmax][nmax];
int carnum[nmax];
int dest;
int capacity;
int min_dst = mint;  //最短路径
int edgenum;
int vnum;
int mark[nmax];

vector<int> tmpvec;
int needSend = 0;
int needBack = 0;

int half_max;
typedef struct path
{
	vector<int> vec; //存储路径
	int needSend;
	int needBack;
	int minPath;
}path;

path tmp_path;
vector<path> pathArray;
void printPath(const vector<int>& vec)
{
	cout << "0";
	for (const auto& p : vec)
	{
		cout << "->" << p;
	}
	cout << " ";
}
void DFS(int src, int dst, int lastBike)
{
	if (dst > min_dst) { return; }
	//如果遍历到了目标结点
	if (src == dest)
	{
		if (dst <= min_dst)
		{
			//如果该节点已经饱和
			if (lastBike + carnum[dest] >= half_max)
			{
				needBack = lastBike + carnum[dest] - half_max;
			}
			//否则补充
			else
			{
				needSend += half_max - (lastBike + carnum[dest]);
			}
			tmp_path.vec = tmpvec;
			tmp_path.needSend = needSend;
			tmp_path.needBack = needBack;
			tmp_path.minPath = dst;
			pathArray.push_back(tmp_path);
			needBack = 0;
			if(lastBike + carnum[dest] < half_max)
				needSend -= half_max - (lastBike + carnum[dest]);
			min_dst = dst;

		}
		return;
	}
	for (int i = 0; i < vnum; ++i)
	{
		if (graph[src][i] != mint && mark[i] == 0)
		{
			mark[i] = 1;
			tmpvec.push_back(i);
			int nextbike = 0;
			if (src != 0)
			{
				//如果该节点已经饱和
				if (lastBike + carnum[src] >= half_max)
				{
					nextbike = lastBike + carnum[src] - half_max;
				}
				//否则补充
				else
				{
					needSend += half_max - (lastBike + carnum[src]);
				}
			}
			DFS(i, dst + graph[src][i],nextbike);
			if (src != 0)
			{
				if (half_max > lastBike + carnum[src])
					needSend -= half_max - (lastBike + carnum[src]);
			}
			else
			{
				needSend = 0;
			}
			tmpvec.pop_back();
			mark[i] = 0;
		}
	}
}

bool cmp(const path& a, const path& b)
{
	if (a.minPath != b.minPath)
		return a.minPath < b.minPath;
	else
	{
		if (a.needSend != b.needSend)
			return a.needSend < b.needSend;
		else
			return a.needBack < b.needBack;
	}
}

int main()
{
	cin >> capacity >> vnum >> dest >> edgenum;
	vnum += 1;
	half_max = capacity / 2;
	for (int i = 0; i < vnum; ++i)
	{
		for (int j = 0; j < vnum; ++j)
		{
			graph[i][j] = mint;
		}
	}
	for (int i = 1; i < vnum; ++i)
	{
		cin >> carnum[i];
	}
	for (int i = 0; i < edgenum; ++i)
	{
		int u, v;
		cin >> u >> v;
		cin >> graph[u][v];
		graph[v][u] = graph[u][v];
	}
	fill(mark, mark + vnum, 0);
	mark[0] = 1;
	DFS(0,0,0);

	sort(pathArray.begin(), pathArray.end(), cmp);
	cout << pathArray[0].needSend << " ";
	printPath(pathArray[0].vec);
	cout << pathArray[0].needBack;
}