Communication System
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 22380 | Accepted: 7953 |
Description
We have received an order from Pizoor
Communications Inc. for a special communication system. The system consists of
several devices. For each device, we are free to choose from several
manufacturers. Same devices from two manufacturers differ in their maximum
bandwidths and prices.
By overall bandwidth (B) we mean the minimum of
the bandwidths of the chosen devices in the communication system and the total
price (P) is the sum of the prices of all chosen devices. Our goal is to choose
a manufacturer for each device to maximize B/P.
Input
The first line of the input file contains a single
integer t (1 ≤ t ≤ 10), the number of test cases, followed by the input data for
each test case. Each test case starts with a line containing a single integer n
(1 ≤ n ≤ 100), the number of devices in the communication system, followed by n
lines in the following format: the i-th line (1 ≤ i ≤ n) starts with mi (1 ≤ mi
≤ 100), the number of manufacturers for the i-th device, followed by mi pairs of
positive integers in the same line, each indicating the bandwidth and the price
of the device respectively, corresponding to a manufacturer.
Output
Your program should produce a single line for each
test case containing a single number which is the maximum possible B/P for the
test case. Round the numbers in the output to 3 digits after decimal
point.
Sample Input
1 3
3 100 25 150 35 80 25
2 120 80 155 40
2 100 100 120 110
Sample Output
0.649
Source
Tehran
2002, First Iran Nationwide Internet Programming Contest
投降的题...
题意：
给出n样零件，及每样零件的都有b[i]个厂生产，有带宽b和价格p两个值。求全部零件(每样一件) min(b)/sum(p) 的最大值。
枚举：
对带宽进行枚举，选出全部零件的最大及最小带宽，然后进行枚举，每个循环选出符合条件的价格的和，得出ans再进行比较，得到最优解。
1 //232K 47MS C++ 1164B 2014-05-07 19:16:42
2 #include<stdio.h>
3 #include<string.h>
4 #include<stdlib.h>
5 struct node{
6 int b;
7 int p;
8 }a[105][105];
9 int b[105];
10 int n;
11 inline int Max(int x,int y)
12 {
13 return x>y?x:y;
14 }
15 inline int Min(int x,int y)
16 {
17 return x<y?x:y;
18 }
19 int main(void)
20 {
21 int t,m;
22 scanf("%d",&t);
23 while(t--)
24 {
25 scanf("%d",&n);
26 int ln=0x7ffffff,rn=0;
27 for(int i=0;i<n;i++){
28 scanf("%d",&b[i]);
29 for(int j=0;j<b[i];j++){
30 scanf("%d%d",&a[i][j].b,&a[i][j].p);
31 ln=Min(ln,a[i][j].b);
32 rn=Max(rn,a[i][j].b);
33 }
34 }
35 double ans=0;
36 for(int i=ln;i<=rn;i++){
37 int sminn=0;
38 for(int j=0;j<n;j++){
39 int minn=0x7ffffff;
40 for(int k=0;k<b[j];k++){
41 if(a[j][k].b>=i && minn>a[j][k].p)
42 minn=a[j][k].p;
43 }
44 sminn+=minn;
45 }
46 ans=ans>(i*1.0/sminn)?ans:(i*1.0/sminn);
47 //printf("*****%lf\n",ans);
48 }
49 printf("%.3lf\n",ans);
50 }
51 return 0;
52 }