PAT 1087

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

题意：求最短路径，以及路数，和路径数量，以及平均最大幸福度

难点：不再是想以前给个数字，可以直接用dij算法，用c++中 map将字符串与数字对应起来比较方便，不然用c写太累了；

#include <stdio.h>
#include <map>
#include <algorithm>
#include <iostream>
#include <vector>
#include <string.h>
#include <string>
#include <limits.h>
using namespace std;
int n,k;
string start;
map<string,int> happy;
map<int ,string>  duiying;
map<string,int> rduiying;
int garph[200][200];
int visit[201]={0};
int path[201]={-1};
int dist[200];
int road[201]={0};
int cc[201]={0};
int renshu[201]={0};

void di()
{
visit[0]=1;
for(int i=1;i<n;i++)
{
if(garph[0][i]!=-1)
{
dist[i]=garph[0][i];
path[i]=0;
road[i]=1;
renshu[i]=1;
}
cc[i]=happy[duiying[i]];
}
dist[0]=0;
while(1)
{
int mindist=1<<30;
int  temp;
for(int i=1;i<n;i++)
{
if(dist[i]<mindist&&visit[i]==0)
{
mindist=dist[i];
temp=i;
}
}
if(mindist==1<<30) break;
visit[temp]=1;
for(int i=1;i<n;i++)
{
if(garph[temp][i]!=-1&&garph[temp][i]+dist[temp]<dist[i])
{
dist[i]=dist[temp]+garph[temp][i];
road[i]=road[temp];
path[i]=temp;
renshu[i]=renshu[temp]+1;
cc[i]=cc[temp]+happy[duiying[i]];
}
else if(garph[temp][i]!=-1&&garph[temp][i]+dist[temp]==dist[i])
{
road[i]+=road[temp];
if(cc[i]<cc[temp]+happy[duiying[i]])
{
cc[i]=cc[temp]+happy[duiying[i]];
path[i]=temp;
}
else if(cc[i]==cc[temp]+happy[duiying[i]])
{
if(renshu[i]>renshu[temp]+1)
renshu[i]=renshu[temp]+1;
}
}
}
}
}
int main()
{
memset(garph,-1,sizeof(garph));
memset(dist,INT_MAX,sizeof(dist));
for(int i=0;i<200;i++)
dist[i]=1<<30;
cin>>n>>k>>start;
duiying[0]=start;
rduiying.insert(make_pair(start,0));
for(int i=1;i<n;i++)
{
string s1;
int hap;
cin>>s1>>hap;
duiying[i]=s1;
rduiying.insert(make_pair(s1,i));
happy.insert(make_pair(s1,hap));
}
for(int i=0;i<k;i++)
{
string c1,c2;
int cost;
cin>>c1>>c2>>cost;
int n1=rduiying[c1];
int n2=rduiying[c2];
garph[n1][n2]=garph[n2][n1]=cost;
}
di();
int kk=rduiying["ROM"];
vector<int> v;
for(int i=kk;i!=0;)
{
v.push_back(i);
i=path[i];
}
//cout<<v.size();
cout<<road[kk]<<" "<<dist[kk]<<" "<<cc[kk]<<" "<<cc[kk]/v.size()<<endl;
cout<<start;
for(int i=v.size()-1;i>=0;i--)
cout<<"->"<<duiying[v[i]];
cout<<endl;
return 0;
}