| Assemble | ||||||
|
||||||
| 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 It will always possible to construct a computer with your budget. |
||||||
| Output | ||||||
|
Per testcase: |
||||||
| 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 | ||||||
| Source | ||||||
| NWERC2007 |
大意：给你一部分电脑零件
每种零件有一定的价格和性能值
每种零件需要一个
电脑的整体性能为所选的所有零件中性能最差的
给你一笔预算和一些零件问你不超过预算的求情况下电脑的最大性能
分析：
这是黄老师家上的一道题
比赛的时候硬是半天没有做出来
其实直接暴力就可以了
枚举每种可能出现的答案 然后判断预算是否超支 对于每个枚举的答案 同一类型如果没有一件物品的性能值大于等于该值说明不符合条件
对于多件符合的物品 选择价格最便宜的即可
代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <map> 6 using namespace std; 7 8 const int maxn = 1005; 9 10 struct point 11 { 12 string s; 13 int a, b; 14 }poin[maxn]; 15 16 int n, sum_b; 17 bool check(int x) 18 { 19 int sum = 0; 20 string s1 = poin[0].s; 21 int st = 0, en = 0;//每一类的最开始的下标和结束的下标 22 for(int i = 0; i < n; i++)//需要注意的是最后一类的判断 23 { 24 if(poin[i].s != s1)//不属于上一类 25 { 26 en = i - 1; 27 bool flag = false; 28 int max_num = 0xffffff; 29 for(int i = st; i <= en; i++)//判断 30 { 31 if(poin[i].b >= x) 32 { 33 flag = true; 34 if(max_num > poin[i].a)//在该类寻找最便宜的即可 35 { 36 max_num = poin[i].a; 37 } 38 } 39 } 40 if(!flag)//该类中没有一件物品的品质大于等于x 41 return false; 42 sum += max_num; 43 st = i; 44 s1 = poin[i].s; 45 } 46 } 47 //printf("*%d ",sum); 48 if(sum <= sum_b) 49 return true; 50 else 51 return false; 52 } 53 54 int main() 55 { 56 int t; 57 //freopen("1511.txt","r",stdin); 58 scanf("%d",&t); 59 while(t--){ 60 scanf("%d %d",&n, &sum_b); 61 string str; 62 int num[maxn] = { 0 };//num数组用于存储需要枚举的值 63 int k = 0; 64 for(int i = 0; i < n; i++) 65 { 66 cin>>poin[i].s>>str>>poin[i].a>>poin[i].b; 67 num[k++] = poin[i].b;//将需要枚举的值加入数组中 68 } 69 poin[n].s = "asdf";//这一步非常重要 wrong一次 70 n++; 71 sort(num, num + k); 72 num[k] = 0; 73 int ans = 0; 74 for(int i = k - 1; i >= 0; i--)//从大到小 75 { 76 if(num[i] == num[i+1]) 77 continue;//枚举过的话就直接忽略 78 if(check(num[i]))//check 用来检测预算是否超支 79 { 80 ans = num[i]; 81 break;//如果满足条件的话直接跳出，因为要求的就是求得最大值 82 } 83 // printf("%d %d ",num[i],check(num[i])); 84 } 85 printf("%d ",ans); 86 } 87 return 0; 88 }