PAT-Advanced Level-1002-All Roads Lead to Rome

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PAT-Advanced Level-1002-All Roads Lead to Rome

题目链接：All Roads Lead to Rome

题目大意：

除起点外每个顶点都有幸运值，寻找一条最短路径到ROM（Rome），若存在多条路径，找出其中幸运值总和最大的路径，若总和相等，找出幸运值平均值最大的路径。

题目分析：

Dijkstra+DFS

详见类同题目的分析：PAT-Advanced Level-1001-Public Bike Management

PAT甲级题解：PAT_ProgramAnswer

代码：

#include<iostream>
#include<map>
#include<string>
#include<vector>
#define MAX_SIZE 210		
#define INF 1<<30
#define Start 1		  //the Start Point of PATH
using namespace std;
//store information of every city
map<string, int> Cities;			
map<int, string> CitiesName;
int Happiness[MAX_SIZE];

int N, K;	//city num ,the num of route
int CitiesMap[MAX_SIZE][MAX_SIZE];		
int RouteCost[MAX_SIZE];		//represent the least cost from start to i
bool Vis[MAX_SIZE];			//Visit Sysbom of every city
vector<int> LeastRoutes[MAX_SIZE];	//store total shortest paths

void Init()
{
	for (int i = 0; i < MAX_SIZE; i++)
		for (int j = 0; j < MAX_SIZE; j++)
			CitiesMap[i][j] = INF;
}

void Dijkstra()
{
	int i, j, k, MIN;
	for (i = 1; i <= N; i++)
	{
		Vis[i] = false;
		RouteCost[i] = CitiesMap[Start][i];
		LeastRoutes[i].push_back(Start);
	}
	Vis[Start] = true;
	for (i = 1; i <= N; i++)
	{
		MIN = INF;
		for (j = 1; j <= N; j++)
		{
			if (!Vis[j] && RouteCost[j] < MIN)
			{
				k = j;
				MIN = RouteCost[j];
			}
		}
		Vis[k] = true;
		for (j = 1; j <= N; j++)
		{
			if (!Vis[j] && RouteCost[k] + CitiesMap[k][j] < RouteCost[j])
			{
				RouteCost[j] = RouteCost[k] + CitiesMap[k][j];
				LeastRoutes[j].clear();
				LeastRoutes[j].push_back(k);
			}
			else if (!Vis[j] && RouteCost[k] + CitiesMap[k][j] == RouteCost[j])
				LeastRoutes[j].push_back(k); 
		}
	}
}
int RouteNum=0;		//the number of shortest paths
int TotalHappiness=0, AverHappiness = 0;		//record the maximum result
vector<int> TmpPath;		//one shortest path in DFS pricess 
vector<int> Res;			//record result shortest path 
void DFS(int v)
{
	if (v == Start)		// produce a compete shortest path 
	{
		RouteNum++;		//add one 
		int Total = 0;
		TmpPath.push_back(v);
		int CityInRoute = TmpPath.size() - 1;
		for (int i =CityInRoute; i >= 0; i--)	
			Total += Happiness[TmpPath[i]];	//cal the sum of happiness 
		if (Total>TotalHappiness)		
		{
			AverHappiness = (int)(Total / CityInRoute);
			TotalHappiness = Total;
			Res = TmpPath;
		}
		else if (Total == TotalHappiness)	
		{
			//choose a path with large averager 
			if ((int)(Total / CityInRoute) > AverHappiness)
			{
				AverHappiness = (int)(Total / CityInRoute);
				Res = TmpPath;
			}
		}
		TmpPath.pop_back();		//backtracking~
		return;
	}
	TmpPath.push_back(v);
	for (int i = 0; i < LeastRoutes[v].size(); i++)
		DFS(LeastRoutes[v][i]);		//traverse all paths below this point
	TmpPath.pop_back();		//backtracking~
}


int main()
{
	Init();
	string CityName;
	int happiness;
	cin >> N >> K >> CityName;
	Cities[CityName] = 1;
	CitiesName[1] = CityName;
	Happiness[1] = 0;
	for (int i = 2; i <= N; i++)
	{
		cin >> CityName >> happiness;
		Cities[CityName] = i;
		Happiness[i] = happiness;
		CitiesName[i] = CityName;
	}
	while (K--)
	{
		string u, v;
		int cost;
		cin >> u >> v >> cost;
		CitiesMap[Cities[u]][Cities[v]] = cost;
		CitiesMap[Cities[v]][Cities[u]] = cost;
	}
	Dijkstra();

	DFS(Cities["ROM"]);
	cout << RouteNum << " " << RouteCost[Cities["ROM"]] << " " << TotalHappiness << " " << AverHappiness << endl;
	for (int i = Res.size() - 1; i >= 0; i--)
	{
		if (i != Res.size() - 1)
			cout << "->";
		cout << CitiesName[Res[i]];
	}
	cout << endl;
	return 0;
}