PTA甲级考试真题练习87——1087 All Roads Lead to Rome

题目

思路

hash字符串+图的最短路径（我使用DFS）

代码

#include <iostream>
#include <string>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
#define nmax 205
#define inf 999999
map<string, int> STOI;
map<int, string> ITOS;
string srcs, tar = "ROM";
int n, m;
int graph[nmax][nmax];
int hapi[nmax];
int visited[nmax];
int _index = 0;
int src, dest;
int mincost = inf;
vector<int> vec;
typedef struct info
{
	int cost;
	int hapi;
	double averhapi;
	vector<int> v;
};
vector <info> vs;

bool cmp(const info& a, const info& b) {
	if (a.cost != b.cost)
		return a.cost < b.cost;
	else
		if (a.hapi != b.hapi)
			return a.hapi > b.hapi;
		else
			return a.averhapi > b.averhapi;
}

int StrToInt(string str) {
	STOI.insert(pair<string, int>(str, _index));
	ITOS.insert(pair<int, string>(_index,str));
	int num = _index;
	++_index;
	return num;
}

void DFS(int cur,int cost,int hapio) {
	if (cur == dest) {
		if (cost <= mincost) {
			info in;
			in.cost = cost;
			in.hapi = hapio;
			in.averhapi = (double)hapio / (vec.size()-1);
			in.v = vec;
			vs.emplace_back(in);
		}
		return;
	}
	for (int i = 0; i < n; ++i) {
		if (visited[i] == 0 && graph[cur][i] != 0) {
			visited[i] = 1;
			vec.emplace_back(i);
			DFS(i, cost + graph[cur][i],hapio+hapi[i]);
			vec.pop_back();
			visited[i] = 0;
		}
	}
}
int main()
{
	cin >> n >> m >> srcs;
	src = StrToInt(srcs);
	hapi[src] = 0;
	for (int i = 0; i < n-1; ++i) {
		string tmp;
		int hapn;
		cin >> tmp >> hapn;
		int _index = StrToInt(tmp);
		if (tmp == tar)
			dest = _index;
		hapi[_index] = hapn;
	}
	memset(graph, 0, sizeof(graph));
	for (int i = 0; i < m; ++i) {
		string u, v;
		int cost;
		cin >> u >> v >> cost;
		int iu = STOI[u];
		int iv = STOI[v];
		graph[iu][iv] = graph[iv][iu] = cost;
	}
	memset(visited, 0, sizeof(visited));
	visited[src] = 1;
	vec.emplace_back(src);
	DFS(src, 0,0);
	sort(vs.begin(), vs.end(), cmp);
	int num = 1;
	int cost = vs[0].cost;
	for (int i = 1; i < vs.size(); ++i) {
		if (vs[i].cost != cost)
			break;
		num++;
	}
	cout << num <<" "<< cost <<" "<< vs[0].hapi <<" "<<(int)vs[0].averhapi << endl;
	cout<<ITOS[vs[0].v[0]];
	for (int i = 1; i < vs[0].v.size(); ++i)
		cout << "->" << ITOS[vs[0].v[i]];
	return 0;
}