pat-1087最短路径

Posted 2020-09-18 jzssuanfa

近期一次pat考试中的最后一题。事实上玩算法这东西就像打魔兽。不能光有思想上的高度，微操必须实打实。就这么个迪杰斯特拉算法。多少教科书上都讲烂了。

可是现场又有多少人是敲对的呢？不能光停留在理解上。必须能用自己的方式表达出来才算过关。

题目:

1087. All Roads Lead to Rome (30)

时间限制

200 ms

内存限制

32000 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

----------------------------------------------------------------------------------------------------------------------------------------------------

题目的意思非常明白，给定图的点和边。起点。终点，求最短路径而且打印。假设有同样的情况，推断总的幸福指数。假设还是同样，推断平均幸福指数。

本题唯一的难点就在于统计同样路径的条数。

我们用数组存储到达每一个点的同样路径的条数。初始值都是1。

实际上就仅仅有两种情况，第一：这个点A被其它点B更新了，那么到达A的同样的最短路径条数就是B点的同样最短路径条数。

第二：起点到A点和B点的长度相等。那么A点的最短路径条数就是A本身的条数加上B点的条数。

// pat-1087.cpp : 定义控制台应用程序的入口点。

//

#include "stdafx.h"

#include"algorithm"
#include"map"
#include"vector"
#include"string"
#include"iostream"
#include"stack"
using namespace std;

#define max 300
#define inf 2100000000//!!!
int dis[max][max]={0};
int n=0;
int visited[max]={0};
int father[max]={0};
int same[max]={0};
int happies[max]={0};
int finalhappies[max]={0};
int precnt[max]={0};
map<int,string> cityname;
map<string,int> citynum;
stack<string> ans;
void dijstrak(int s)
{  
	same[0]=1;
	visited[s]=1;
	for(int i=0;i<n;i++)
	{
		 int min=inf;
		 int mark=-1;
	     for(int j=0;j<n;j++)
		 {//find min
		     if(visited[j]==0&&dis[s][j]<min)
			 {
			    min=dis[s][j];
			    mark=j;
			 }
		 }
		 if(mark==-1)
			 return ;
		 visited[mark]=1;//marked
		 for(int k=0;k<n;k++)
		 {//updata
			 if(visited[k]==1)
				 continue;
			 int raw=dis[s][k];
			 int another=dis[s][mark]+dis[mark][k];
		    if(raw>another)
			{   same[k]=same[mark];
		        dis[s][k]=dis[s][mark]+dis[mark][k];
				father[k]=mark;
				finalhappies[k]=finalhappies[mark]+happies[k];
				precnt[k]=precnt[mark]+1;
			}
			else if(raw==another)
			{
				same[k]+=same[mark];
				if(finalhappies[k]<finalhappies[mark]+happies[k])
				{
					father[k]=mark;
					finalhappies[k]=finalhappies[mark]+happies[k];
					precnt[k]=precnt[mark]+1;
				}
				else if(finalhappies[k]==finalhappies[mark]+happies[k])
				{
				   if(precnt[k]>(precnt[mark]+1))
				   {
					   father[k]=mark;
					   precnt[k]=precnt[mark]+1;
				   }
				}
			}

			}
				
		 }
	}

int main()
{
	for(int i=0;i<max;i++)
	    for(int j=0;j<max;j++){
			dis[i][j]=inf;
			dis[j][i]=inf;
			dis[i][i]=inf;
			dis[j][j]=inf;
	}
   int k=0;
   string s;
   cin>>n>>k>>s;
   cityname[0]=s;
   string citystr;
   int happy=0;
   for(int i=1;i<n;i++)
   {
	cin>>citystr>>happy;
	cityname[i]=citystr;
	citynum[citystr]=i;
	happies[i]=happy;
	finalhappies[i]=happy;
	precnt[i]=1;
	same[i]=1;
   }
   string city1,city2;
   int cost=0;
   for(int i=0;i<k;i++){
     cin>>city1>>city2>>cost;
	 int j=citynum[city1];
	 int b=citynum[city2];
	 dis[j][b]=cost;
	 dis[b][j]=cost;
   }

   dijstrak(0);   

   int romanum=citynum["ROM"];
   int index=romanum;
   string path;
   path+=s;//"HZH"
   ans.push("ROM");
   while(father[index]!=0)
   {
	 string pathcity=cityname[father[index]];
	 ans.push(pathcity);
	 index=father[index];
   }
   while(!ans.empty()){
	   path+="->";
	   path+=ans.top();
	   ans.pop();
   }
   cout<<same[romanum]<<" "<<dis[0][romanum]<<" "<<finalhappies[romanum]<<" "<<finalhappies[romanum]/precnt[romanum]<<endl;
   cout<<path<<endl;
  return 0;
}

提交的时候把预编译头

#include "stdafx.h"

去掉就可以。