Pat(Advanced Level)Practice--1055(The World's Richest)

Pat1055代码

题目描述：

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

超时代码：

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#define MAX 205

using namespace std;

typedef struct Richer
{
	char name[10];
	int age;
	int wealth;
}Richer;

bool cmp(const Richer &l,const Richer &r)
{
	if(l.wealth==r.wealth)
		return strcmp(l.name,r.name)>0;
	else
		return l.wealth<r.wealth;
}

int main(int argc,char *argv[])
{
	int n,k;
	int i,j;
	int m,Amin,Amax;
	vector<Richer> v[MAX];
	scanf("%d%d",&n,&k);
	getchar();
	for(i=0;i<n;i++)
	{
		Richer temp;
		scanf("%s %d %d",temp.name,&temp.age,&temp.wealth);
		getchar();
		v[temp.age].push_back(temp);
	}
	for(i=0;i<MAX;i++)
		sort(v[i].begin(),v[i].end(),cmp);
	for(i=0;i<k;i++)
	{
		vector<Richer> t[MAX];
		for(j=0;j<MAX;j++)
			t[j]=v[j];
		scanf("%d%d%d",&m,&Amin,&Amax);
		int total=0;
		for(j=Amin;j<=Amax;j++)
			total+=t[j].size();
		if(total<m)
			m=total;
		printf("Case #%d:\n",i+1);
		if(m==0)
		{
			printf("None\n");
			continue;
		}
		while(m)
		{
			int max=-1000001;
			int index;
			int end;
			for(j=Amin;j<=Amax;j++)
			{
				int end=t[j].size();
				if(end!=0&&(t[j][end-1]).wealth>max)
				{
					max=(t[j][end-1]).wealth;
					index=j;
				}
			}
			end=t[index].size();
			printf("%s %d %d\n",t[index][end-1].name,
			                t[index][end-1].age,t[index][end-1].wealth);
			t[index].pop_back();
			m--;
		}
	}

	return 0;
}

估计vector的复制需要大量的时间，有2个case超时。

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAX 100005

using namespace std;

typedef struct Richer
{
	char name[10];
	int age;
	int wealth;
}Richer;

bool cmp(const Richer &l,const Richer &r)
{
	if(l.wealth!=r.wealth)
		return l.wealth>r.wealth;
	else if(l.age!=r.age)
		return l.age<r.age;
	else
		return strcmp(l.name,r.name)<0;
}

int main(int argc,char *argv[])
{
	Richer r[MAX];
	int Agecount[205];//统计每个年龄的人数
	int Sorted[MAX];//按财富大小存储在数组中的下标
	int n,k;
	int i,j;
	int t=0;
	int m,Amin,Amax;
	scanf("%d%d",&n,&k);
	for(i=0;i<n;i++)
		scanf("%s %d %d",r[i].name,&r[i].age,&r[i].wealth);
	sort(r,r+n,cmp);
	memset(Agecount,0,sizeof(Agecount));
	t=0;
	for(i=0;i<n;i++)
	{
		Agecount[r[i].age]++;
		if(Agecount[r[i].age]<101)//应为M的值最大为100，所以如果某个年龄
			Sorted[t++]=i;  //段的人数超过100，后面的需要忽略
	}
	for(i=1;i<=k;i++)
	{
		int num=0;
		printf("Case #%d:\n",i);
		scanf("%d%d%d",&m,&Amin,&Amax);
		for(j=0;j<t&&num<m;j++)
		{
			int age=r[Sorted[j]].age;
			if(age>=Amin&&age<=Amax)
			{
				printf("%s %d %d\n",r[Sorted[j]].name,r[Sorted[j]].age,
				r[Sorted[j]].wealth);
				num++;
			}
		}
		if(num==0)
			printf("None\n");
	}

	return 0;
}

需要将该年龄出现次数超过100的过滤掉，否则会超时。