pat 1087

青旅半醒 2022-08-21 06:45 103阅读 0赞

# 1087. All Roads Lead to Rome (30) #

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness.

**Input Specification:**

Each input file contains one test case. For each case, the first line contains 2 positive integers N (2<=N<=200), the number of cities, and K, the total number of routes between pairs of cities; followed by the name of the starting city. The next N-1 lines each gives the name of a city and an integer that represents the happiness one can gain from that city, except the starting city. Then K lines follow, each describes a route between two cities in the format "City1 City2 Cost". Here the name of a city is a string of 3 capital English letters, and the destination is always ROM which represents Rome.

**Output Specification:**

For each test case, we are supposed to find the route with the least cost. If such a route is not unique, the one with the maximum happiness will be recommended. If such a route is still not unique, then we output the one with the maximum average happiness -- it is guaranteed by the judge that such a solution exists and is unique.

Hence in the first line of output, you must print 4 numbers: the number of different routes with the least cost, the cost, the happiness, and the average happiness (take the integer part only) of the recommended route. Then in the next line, you are supposed to print the route in the format "City1->City2->...->ROM".

**Sample Input:**

6 7 HZH
    ROM 100
    PKN 40
    GDN 55
    PRS 95
    BLN 80
    ROM GDN 1
    BLN ROM 1
    HZH PKN 1
    PRS ROM 2
    BLN HZH 2
    PKN GDN 1
    HZH PRS 1

**Sample Output:**

3 3 195 97
    HZH->PRS->ROM

图论题目，不算难，但为啥我的代码提交后出现段错误，编译器是g++,本地vs测试没问题啊，why

#include<iostream>
    #include<vector>
    #include<string>
    #include<map>
    using namespace std;
    int cal_cost(vector<string>&path, map<pair<string, string>, int>&cost)
    {
    	int co = 0;
    	for (int i = 1; i<path.size(); i++)
    		co += cost[pair<string, string>(path[i - 1], path[i])];
    	return co;
    }
    int cal_hap(vector<string>&path, map<string, int>&happi)
    {
    	int happy = 0;
    	for (int i = 1; i<path.size(); i++)
    		happy += happi[path[i]];
    	return happy;
    }
    int main()
    {
    	int N = 6, K = 7;
    	map<string, vector<string> >graph;
    	map<string, int>happi;
    	map<pair<string, string>, int>cost;
    	string str;
    	/*cin >> N>>K>>str;
    	for (int i = 0; i<N - 1; i++)
    	{
    		int ha;
    		cin >> str >> ha;
    		happi[str] = ha;
    	}
    	for (int i = 0; i<K; i++)
    	{
    		string str1; int cos;
    		cin >> str>>str1>>cos;
    		graph[str].push_back(str1); 
    		pair<string, string>po = pair<string, string>(str, str1);
    		cost[po] = cos;
    	}*/
    	
    	happi["ROM"] = 100;
    	happi["PKN"] = 40;
    	happi["GDN"] = 55;
    	happi["PRS"] = 95;
    	happi["BLN"] = 80;
    	cost[pair<string, string>("ROM", "GDN")] = 1;
    	graph["ROM"].push_back("GDN");
    	cost[pair<string, string>("BLN", "ROM")] = 1;
    	graph["BLN"].push_back("ROM");
    	cost[pair<string, string>("HZH", "PKN")] = 1;
    	graph["HZH"].push_back("PKN");
    	cost[pair<string, string>("PRS", "ROM")] = 2;
    	graph["PRS"].push_back("ROM");
    	cost[pair<string, string>("BLN", "HZH")] = 2;
    	graph["BLN"].push_back("HZH");
    	cost[pair<string, string>("PKN", "GDN")] = 1;
    	graph["PKN"].push_back("GDN");
    	cost[pair<string, string>("HZH", "PRS")] = 1;
    	graph["HZH"].push_back("PRS");
    	vector<string>que;
    	int minco = 10000000;
    	que.push_back("HZH");
    	vector<vector<string> >candi;
    	while (!que.empty())
    	{
    		if (que.back() == "ROM")
    		{
    			int curco = cal_cost(que, cost);
    			if (curco == minco)
    				candi.push_back(que);
    			else if (curco<minco)
    			{
    				minco = curco;
    				candi.clear();
    				candi.push_back(que);
    			}
    			que.pop_back();
    		}
    		else
    		{
    			while (!que.empty() && graph.find(que.back())==graph.end())
    				que.pop_back();
    			if (!que.empty())
    			{
    				que.push_back(graph[que.back()].back());
    				graph[que[que.size() - 2]].pop_back();
    				if (graph[que[que.size() - 2]].empty())
    					graph.erase(graph.find(que[que.size() - 2]));
    			}
    		}
    	}
    	int routenum = candi.size();
    	int mosthap = cal_hap(candi[0], happi);
    	que = candi[0];
    	int avghap;
    	for (int i = 1; i<candi.size(); i++)
    	{
    		int hh = cal_hap(candi[i], happi);
    		if (hh>mosthap)
    		{
    			mosthap = hh;
    			que = candi[i];
    		}
    	}
    	avghap = mosthap / (que.size()-1);
    	cout << candi.size() << " " << minco << " " << mosthap << " " << avghap << endl;
    	for (int i = 0; i<que.size() - 1; i++)
    		cout << que[i] << "->";
    	cout << que.back();
    	return 0;
    }