hdoj 2333 Assemble

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. 

InputOn 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. 
OutputPer 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

 1 /*
 2 解題思路:
 3       1. 求解這類"最小值最大"的問題通常方法"二分法". 肯定最大和最小的quality, 經過二分
 4          枚舉, 判斷是否知足最小价錢不超過b價格. 問題天然求解.
 5       2. poj這題數據有點BT, 好屢次TLE, 合理使用STL才行. 用map查找配件類型, 875MS時間
 6          不高, 後來網上發現根據quality排序能夠在判斷是否知足時候提速, 好方法.
 7 */
 8 #include <cstdio>
 9 #include <iostream>
10 #include <cstring>
11 #include <string>
12 #include <vector>
13 #include <map>
14 using namespace std;
15 #define MAX 1005
16 const int INF = (1<<29);
17 
18 struct node{
19     int price, qu;
20 }good[MAX][MAX];
21 
22 int n, b;
23 map pos;
24 int count[MAX];
25 int num;
26 
27 inline int max(int a, int b){
28     return a > b ? a : b;
29 }
30 
31 inline int min(int a, int b){
32     return a < b ? a : b;
33 }
34 
35 inline int find(string str){
36     if( !pos.count(str) ) pos[str] = num++;
37     return pos[str];
38 }
39 
40 inline bool judge(int qu){
41     int result = 0;
42     for(int i = 0; i < num; ++i){
43         int minOne = b+1;
44         for(int j = 0; j < count[i]; ++j){
45             if( good[i][j].qu >= qu )
46                 minOne = min(minOne, good[i][j].price);
47         }
48         result += minOne;
49         if(result > b) return false;
50     }
51     return true;
52 }
53 
54 int main(){
55     int caseNum;
56     scanf("%d", &caseNum);
57     while(caseNum--){
58         scanf("%d %d", &n, &b);
59         num = 0;
60 
61         memset(count, 0, sizeof(count));
62         pos.clear();
63         int left = INF, right = 0;
64         int price, qu;
65         char name[22], type[22];
66         for(int i = 0; i < n; ++i){
67             scanf("%s %s %d %d", type, name, &price, &qu);
68             right = max(right, qu);
69             left = min(left, qu);
70             int temp = find(type);
71             good[temp][count[temp]].price = price;
72             good[temp][count[temp]++].qu = qu;
73         }
74 
75         while(left <= right){
76             int mid = (left+right)>>1;
77             if( judge(mid) ) left = mid+1;
78             else right = mid-1;
79         }
80         printf("%d\n", right);
81     }
82     return 0;
83 }