poj 1018 Communication System (枚举)

 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 }