PAT A1062:Talent and Virtue之排序策略

题目描述

1062 Talent and Virtue (25分)
About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people’s talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a sage(圣人); being less excellent but with one’s virtue outweighs talent can be called a nobleman(君子); being good in neither is a fool man(愚人);
yet a fool man is better than a small man(小人) who prefers talent than virtue.

Now given the grades of talent and virtue of a group of people, you are supposed to rank them according to Sima Guang’s theory.

Input Specification:
Each input file contains one test case. Each case first gives 3 positive integers in a line: N (≤10^5 ), the total number of people to be ranked; L (≥60), the lower bound of the qualified grades – that is, only the ones whose grades of talent and virtue are both not below this line will be ranked; and H (100), the higher line of qualification – that is, those with both grades not below this line are considered as the sages, and will be ranked in non-increasing order according to their total grades. Those with talent grades below H but virtue grades not are cosidered as the noblemen, and are also ranked in non-increasing order according to their total grades, but they are listed after the sages.
Those with both grades below H, but with virtue not lower than talent are considered as the fool men.
They are ranked in the same way but after the noblemen. The rest of people whose grades both pass the L line are ranked after the fool men.

Then N lines follow, each gives the information of a person in the format:

ID_Number Virtue_Grade Talent_Grade

where ID_Number is an 8-digit number, and both grades are integers in [0, 100]. All the numbers are separated by a space.

Output Specification:
The first line of output must give M (≤N), the total number of people that are actually ranked. Then M lines follow, each gives the information of a person in the same format as the input, according to the ranking rules. If there is a tie of the total grade, they must be ranked with respect to their virtue grades in non-increasing order. If there is still a tie, then output in increasing order of their ID’s.

Sample Input:

14 60 80
10000001 64 90
10000002 90 60
10000011 85 80
10000003 85 80
10000004 80 85
10000005 82 77
10000006 83 76
10000007 90 78
10000008 75 79
10000009 59 90
10000010 88 45
10000012 80 100
10000013 90 99
10000014 66 60

Sample Output:

12
10000013 90 99
10000012 80 100
10000003 85 80
10000011 85 80
10000004 80 85
10000007 90 78
10000006 83 76
10000005 82 77
10000002 90 60
10000014 66 60
10000008 75 79
10000001 64 90

求解思路
  • 题意:司马光的评价理论——圣人:talent和virtue均优秀的人;君子:talent和virtue没那么优秀但是virtue不低于talent的人;愚人:talent和virtue均不优秀的人;小人:比愚人等级更低且virtue低于talent的人。
  • 给定n个人及其及格分数L和优秀分数H,对于每个人给出id,virtue分数和talent分数。定义四个vector数组分别用于存储合格情况下圣人、君子、愚人、其他情况。之后按照圣人、君子、愚人、其他情况输出。每种类中排序策略为:总分数不相等的情况下按照总分数非递增排序;总分数相等virtue分数不相等的情况下按照virtue分数非递增排序;如果总分数和virtue分数均相等的情况下,按照id升序排列。
代码实现(AC)
#includecstdio
#includecstring
#includevector
#includealgorithm
using namespace std;
struct Person{
	char id[10];
	int virtue;
	int talent;
};
vectorPersonp1,p2,p3,p4;
int n,l,h;
bool cmp(Person a,Person b)
{
	if((a.virtue+a.talent)!=(b.virtue+b.talent))	return (a.virtue+a.talent)=(b.virtue+b.talent);
	else
	{
		if(a.virtue!=b.virtue)	return a.virtue=b.virtue;
		else return strcmp(a.id,b.id)0;
	}
}
void init()
{
	scanf("%d %d %d",n,l,h);
	for(int i=0;in;i++)
	{
		char str[10];
		int v,t;
		Person p;
		scanf("%s %d %d",str,v,t);
		if(v=lt=l)
		{
			if(v=ht=h)	//圣人 
			{
				strcpy(p.id,str);
				p.talent=t;
				p.virtue=v;
				p1.push_back(p); 
			}
			else if(v=hth)	//君子
			{
				strcpy(p.id,str);
				p.talent=t;
				p.virtue=v;
				p2.push_back(p); 
			}
			else if(vhthv=t) //愚人
			{
				strcpy(p.id,str);
				p.talent=t;
				p.virtue=v;
				p3.push_back(p); 
			}
			else	//其他 
			{
				strcpy(p.id,str);
				p.talent=t;
				p.virtue=v;
				p4.push_back(p); 
			} 
		}
	}
}
void solve()
{
	init();
	int len=p1.size()+p2.size()+p3.size()+p4.size();
	printf("%d\n",len);
	sort(p1.begin(),p1.end(),cmp);
	sort(p2.begin(),p2.end(),cmp);
	sort(p3.begin(),p3.end(),cmp);
	sort(p4.begin(),p4.end(),cmp);
	for(int i=0;ip1.size();i++)
	{
		printf("%s %d %d\n",p1[i].id,p1[i].virtue,p1[i].talent);
	}
	for(int i=0;ip2.size();i++)
	{
		printf("%s %d %d\n",p2[i].id,p2[i].virtue,p2[i].talent);
	}
	for(int i=0;ip3.size();i++)
	{
		printf("%s %d %d\n",p3[i].id,p3[i].virtue,p3[i].talent);
	}
	for(int i=0;ip4.size();i++)
	{
		printf("%s %d %d\n",p4[i].id,p4[i].virtue,p4[i].talent);
	}
}
int main()
{
	solve();
	return 0;
}
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