[图的遍历&多标准] 1087. All Roads Lead to Rome (30)

1087. All Roads Lead to Rome (30)

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

分析：这是一道图的遍历题目，但是涉及到多个判断标准。首先，路径长度(花费)最小。其次，获得的happy值最大。再次，happy的平均值最大(不包括开始的城市).虽然增加了判断标准，但是做法还是一样的。该题基于图的深度遍历DFS来做，每次遍历到目的节点时更新判断标准。另外，图的DFS与数的DFS不同的地方在于，图的DFS需要增加一个vis数组用于表示某个节点是否访问过，而树不需要，因为树是不含环的。另外，该题给的是字符串表示的节点，我们可以用map来实现字符串和int型之间的映射，方便编写代码。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;

const int maxn=210;
const int INF=1e9;

int g[maxn][maxn];
int happy[maxn];
int vis[maxn]={false};

map<string ,int> str2int;
map<int,string> int2str;

int st,ed;
vector<int> tmppath,path;
int min_cost=INF;
int max_happy=0;
int avg_happy=0;
int lest_num=0;

int n;

void dfs(int s,int cost,int hy)
{
    vis[s]=true;
    if(s==ed)
    {
        if(cost<min_cost)
        {
            min_cost=cost;
            path=tmppath;
            lest_num=1;
            max_happy=hy;
            avg_happy=hy/(path.size()-1);
        }
        else if(cost==min_cost)
        {
            lest_num+=1;
            if(hy>max_happy)
            {
                max_happy=hy;
                path=tmppath;
                avg_happy=hy/(path.size()-1);
            }
            else if(hy==max_happy)
            {
                if(tmppath.size()<path.size()&&tmppath.size()>1)
                {
                    path=tmppath;
                    int down=path.size()-1;
                    avg_happy=hy/down;
                }
            }
        }
        return ;
    }
    for(int v=0;v<n;v++)
    {
        if(vis[v]==false&&g[s][v]!=INF)
        {
            tmppath.push_back(v);
            
            dfs(v,cost+g[s][v],hy+happy[v]);
            vis[v]=false;
            tmppath.pop_back();
        }
    }
}

int main()
{
    fill(g[0],g[0]+maxn*maxn,INF);
    int k;
    string begin;
    cin>>n>>k>>begin;
    str2int.insert(make_pair(begin,0));
    int2str.insert(make_pair(0,begin));
    for(int i=1;i<n;i++)
    {
        string str;
        int h;
        cin>>str>>h;
        if(str=="ROM") ed=i;
        str2int.insert(make_pair(str,i));
        int2str.insert(make_pair(i,str));
        happy[i]=h;
    }
    for(int i=0;i<k;i++)
    {
        string u,v;
        int cost;
        cin>>u>>v>>cost;
        int uu,vv;
        uu=str2int[u];
        vv=str2int[v];
        g[uu][vv]=g[vv][uu]=cost;
    }
    tmppath.push_back(0);
    dfs(0,0,0);
    cout<<lest_num<<" "<<min_cost<<" "<<max_happy<<" "<<avg_happy<<endl;
    
    for(int i=0;i<path.size();i++)
    {
        if(i>0) cout<<"->";
        cout<<int2str[path[i]];
    }
}