南邮 OJ 1499 Assemble

Assemble

时间限制(普通/Java) :  3000 MS/ 9000 MS          运行内存限制 : 65536 KByte
总提交 : 23            测试通过 : 10 

比赛描述

Recently your team noticed that the computer you use to practice for programming contests is not good enough anymore. Therefore, you decide to buy a new computer.

To make the ideal computer for your needs, you decide to buy separate components and assemble the computer yourself. You need to buy exactly one of each type of component.

The problem is which components to buy. As you all know, the quality of a computer is equal to the quality of its weakest component. Therefore, you want to maximize the quality of the component with the lowest quality, while not exceeding your budget.

输入

On the first line one positive number: the number of testcases, at most 100. After that per testcase:

• One line with two integers: 1 ≤ n ≤ 1 000 , the number of available components and 1 ≤ b ≤ 1 000 000 000 , your budget.

• n lines in the following format: “type name price quality”, where type is a string with the type of the component, name is a string with the unique name of the component,price is an integer (0 ≤ price ≤ 1 000 000 ) which represents the price of the component and quality is an integer (0 ≤ quality ≤ 1 000 000 000 ) which represents the quality of the component (higher is better). The strings contain only letters, digits and underscores and have a maximal length of 20 characters.

输出

Per testcase:

• One line with one integer: the maximal possible quality.

样例输入

1
18 800
processor 3500_MHz 66 5
processor 4200_MHz 103 7
processor 5000_MHz 156 9
processor 6000_MHz 219 12
memory 1_GB 35 3
memory 2_GB 88 6
memory 4_GB 170 12
mainbord all_onboard 52 10
harddisk 250_GB 54 10
harddisk 500_FB 99 12
casing midi 36 10
monitor 17_inch 157 5
monitor 19_inch 175 7
monitor 20_inch 210 9
monitor 22_inch 293 12
mouse cordless_optical 18 12
mouse microsoft 30 9
keyboard office 4 10

样例输出

9

题目来源

NWERC 2007

#include<iostream>
#include<algorithm>
#include<set>
#define TYPE_LEN 21
#define MAX_N 1001
using namespace std;

struct component{
	char type[TYPE_LEN];
	int price;
	int quality;
	int typeNo;
}c[MAX_N];

//按照名字字典序排序，名字相同的话，按照价格从低到高排序
bool cmp1(const component &c1,const component &c2){
	int temp=strcmp(c1.type,c2.type);
	if(temp){
		return temp<0;
	}
	return c1.price < c2.price;
}

//按照质量升序排序，质量相同的话，按照价格从高到低排序,☆要将价格高的先替换出去
struct cmp2{
	bool operator()(const component &c1,const component &c2)const{
		if(c1.quality != c2.quality){
			return c1.quality < c2.quality;
		}
		return c1.price > c2.price;
	}
};

int beg[MAX_N],end[MAX_N];

int main(){
//	freopen("test.txt","r",stdin);
	int t,n,b,maxTypeNo,i;
	multiset<component,cmp2> s;
	multiset<component,cmp2>::iterator it;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&b);
		getchar();
		for(i=0;i<n;i++){
			scanf("%s%*s%d%d",c[i].type,&c[i].price,&c[i].quality);
		}
		sort(c,c+n,cmp1);
		maxTypeNo = 0;
		i = 0;
		while(i<n){
			beg[maxTypeNo] = i;//此类型的最小下标
			c[i].typeNo = maxTypeNo;
			while(strcmp(c[i].type,c[i+1].type)==0){
				i++;
				c[i].typeNo = maxTypeNo;
			}
			i++;
			end[maxTypeNo] = i;//此类型的最大下标的下一个位置
			maxTypeNo++;
		}
		s.clear();
		int cb = 0;				//当剩余的前
		for(i=0;i<maxTypeNo;i++){
			s.insert(c[beg[i]]);
			cb += c[beg[i]].price;
			beg[i]++;
		}
		int r;					//结果
		while(1){
			it = s.begin();
			int x = (*it).typeNo;	//当前最低质量的编号
			r = (*it).quality;		//当前的最低质量
			while(beg[x]<end[x] && (*it).quality>=c[beg[x]].quality){
				beg[x]++;
			}
			if(beg[x]==end[x]){
				break;						//已经到达这一类的最高质量了
			}
			cb -= (*it).price;
			s.erase(it);
			cb += c[beg[x]].price;
			if(cb > b){
				break;
			}
			component temp;
			temp.typeNo = x;
			temp.price = c[beg[x]].price;
			temp.quality = c[beg[x]].quality;
			s.insert(temp);
			beg[x]++;
		}
		printf("%d\n",r);
	}
}