| 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
#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,&b);
for(int i=1;i<=n;i++)
{
scanf("%s %s %d %d
",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
",anslist[ans]);
}
return 0;
}附书上解释：
【分析】
在《入门经典》一书中，我们曾提到过，解决“最小值最大”的常用方法是二分答案。假设答案为x，如何判断这个x是最小还是最大呢？删除品质因子小于x的所有配件，如果可以组装出一台不超过b元的电脑，那么标准答案ans≥x，否则ans＜x。
如何判断是否可以组装出满足预算约束的电脑呢？很简单，每一类配件选择最便宜的一个即可。如果这样选都还超预算的话，就不可能有解了。代码如下。