【二分答案】 【POJ3497】 【Northwestern Europe 2007】 Assemble 組裝電腦

Assemble

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3171		Accepted: 1013

Description

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.

Input

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.

Output

Per testcase:

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

Sample Input

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

Sample Output

9

好久沒作的二分答案的題目了， 很好的二分答案練手題目。

這種最大中最小，最小中最大通常都是二分答案

而後判斷答案的正確與否

代碼以下：

#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef struct node
{
	char name[30];
	char type[30];
	int price;
	int quality;
	int TYPE;
}node;
node thing[1010];
int flag[1010];
int anslist[1010]; 
int n,b,cnt,ans;
int cmp(const void *i,const void *j)
{
	node *ii=(node *)i,*jj=(node *)j;
	return strcmp(ii->type,jj->type);
}
int cmp3(const void *i,const void *j)
{
	return *(int *)i-*(int *)j;
}
int OK(int std)
{
	int sum=0,PRICE; 
	for(int i=1;i<=cnt;i++)
	 {
	 	PRICE=1000001;
	 for(int j=flag[i];j<flag[i+1];j++)
	   		if(thing[j].quality>=std)
	   		 if(thing[j].price<PRICE) PRICE=thing[j].price;
	    if(PRICE==1000001) return 0;
		else sum=sum+PRICE; 
	}
	if(sum<=b) return 1;
	else return 0;
}
int getans()
{
	int i,j,m;
	i=1;j=n;
	while(i<j)
	{
		m=(i+j)/2+1;
		if(OK(anslist[m])) i=m;
		else j=m-1;  
	}	
	ans=i;
}
int main()
{
//	freopen("a.in","r",stdin);
//	freopen("a.out","w",stdout);
	int T;
	scanf("%d",&T);
	while(T--)
	{
		cnt=1;
		scanf("%d%d\n",&n,&b);
		for(int i=1;i<=n;i++)
		{
			scanf("%s %s %d %d\n",thing[i].type,thing[i].name,&thing[i].price,&thing[i].quality);	
			anslist[i]=thing[i].quality;
		}
		qsort(thing+1,n,sizeof(thing[1]),cmp);
		qsort(anslist+1,n,sizeof(anslist[1]),cmp3); 
		flag[cnt]=1;
		for(int i=2;i<=n;i++)
		{
			
			if(strcmp(thing[i].type,thing[i-1].type)==0) thing[i-1].TYPE=cnt;
			else 
			{
				thing[i-1].TYPE=cnt;
				cnt++;
				flag[cnt]=i;
			}
		}
		flag[cnt+1]=n+1;
			thing[n].TYPE=cnt;
		getans();	
		printf("%d\n",anslist[ans]);
	}
	return 0;
}

附書上解釋：

【分析】

在《入門經典》一書中，咱們曾提到過，解決「最小值最大」的經常使用方法是二分答案。假設答案爲x，如何判斷這個x是最小仍是最大呢？刪除品質因子小於x的全部配件，若是能夠組裝出一臺不超過b元的電腦，那麼標準答案ans≥x，不然ans＜x。

如何判斷是否能夠組裝出知足預算約束的電腦呢？很簡單，每一類配件選擇最便宜的一個便可。若是這樣選都還超預算的話，就不可能有解了。代碼以下。