1055 The World's Richest (25 分) PAT 甲级排序题解决超时问题

雨点打透心脏的1/2处 2024-02-19 21:29 24阅读 0赞

## 1055 The World's Richest (25 分) ##

Forbes magazine publishes every year its list of billionaires based on the annual ranking of the world's wealthiest people. Now you are supposed to simulate this job, but concentrate only on the people in a certain range of ages. That is, given the net worths of N people, you must find the M richest people in a given range of their ages.

#### Input Specification: ####

Each input file contains one test case. For each case, the first line contains 2 positive integers: N (≤105) - the total number of people, and K (≤103) - the number of queries. Then N lines follow, each contains the name (string of no more than 8 characters without space), age (integer in (0, 200\]), and the net worth (integer in \[−106,106\]) of a person. Finally there are K lines of queries, each contains three positive integers: M (≤100) - the maximum number of outputs, and \[`Amin`, `Amax`\] which are the range of ages. All the numbers in a line are separated by a space.

#### Output Specification: ####

For each query, first print in a line `Case #X:` where `X` is the query number starting from 1. Then output the M richest people with their ages in the range \[`Amin`, `Amax`\]. Each person's information occupies a line, in the format

Name Age Net_Worth

The outputs must be in non-increasing order of the net worths. In case there are equal worths, it must be in non-decreasing order of the ages. If both worths and ages are the same, then the output must be in non-decreasing alphabetical order of the names. It is guaranteed that there is no two persons share all the same of the three pieces of information. In case no one is found, output `None`.

#### Sample Input: ####

12 4
    Zoe_Bill 35 2333
    Bob_Volk 24 5888
    Anny_Cin 95 999999
    Williams 30 -22
    Cindy 76 76000
    Alice 18 88888
    Joe_Mike 32 3222
    Michael 5 300000
    Rosemary 40 5888
    Dobby 24 5888
    Billy 24 5888
    Nobody 5 0
    4 15 45
    4 30 35
    4 5 95
    1 45 50

#### Sample Output: ####

Case #1:
    Alice 18 88888
    Billy 24 5888
    Bob_Volk 24 5888
    Dobby 24 5888
    Case #2:
    Joe_Mike 32 3222
    Zoe_Bill 35 2333
    Williams 30 -22
    Case #3:
    Anny_Cin 95 999999
    Michael 5 300000
    Alice 18 88888
    Cindy 76 76000
    Case #4:
    None

这题，我刚开始超时，后来，所有输入输出都用scanf和printf。

又把s.size()都复制给一个变量。

才过了第二个测试点。

这里要说说用scanf输入string类型以及printf输出string类型。

#include <stdio.h>
    #include <string>
    using namespace std;
    int main()
    {
    	string a;
    	a.resize(100); //需要预先分配空间 这点很重要，否则报错
    	scanf("%s", &a[0]);
    	//puts(a.c_str());
    	printf("%s",a.c_str());
    	return 0;
    }

代码：

#include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<vector>
    #include<algorithm>
    using namespace std;
    struct node{
    	string name;
    	int age,worth;
    };
    int cmp(node a,node b)
    {
    	if(a.worth!=b.worth)
    	return a.worth>b.worth;
    	else
    	{
    		if(a.age!=b.age)
    		return a.age<b.age;
    		else
    		return a.name<b.name;
    	}
    }
    int main()
    {
    	int i,j,k,n;
    	int t,min_age,max_age;
    	vector<node> s;
    	node temp;
    	temp.name.resize(9);
    	cin>>n>>k;
    	for(i=0;i<n;i++)
    	{
    		scanf("%s %d %d",&temp.name[0],&temp.age,&temp.worth);
    		s.push_back(temp);
    	}
    	sort(s.begin(),s.end(),cmp);
    	for(i=1;i<=k;i++)
    	{
    		cin>>t>>min_age>>max_age;
    		printf("Case #%d:\n",i);
    		int f=t,len=s.size();
    		for(j=0;j<len;j++)
    		{
    			if(t==0)
    			break;
    			if(s[j].age>=min_age&&s[j].age<=max_age&&t)
    			{
    				printf("%s %d %d\n",s[j].name.c_str(),s[j].age,s[j].worth);
    				t--;
    			}
    			
    		}
    		if(t==f)
    		printf("None\n");
    	}
    	
    	return 0;
    }