地址：http://poj.org/problem?id=1434
题目：Fill the Cisterns!
Fill the Cisterns!
| Time Limit: 5000MS | Memory Limit: 10000K |
| Total Submissions: 4131 | Accepted: 1360 |
Description
During the next century certain regions on earth will experience severe water shortages. The old town of Uqbar has already started to prepare itself for the worst. Recently they created a network of pipes connecting the cisterns that distribute water in each neighbourhood, making it easier to fill them at once from a single source of water. But in case of water shortage the cisterns above a certain level will be empty since the water will to the cisterns below.
You have been asked to write a program to compute the level to which cisterns will be lled with a certain volume of water, given the dimensions and position of each cistern. To simplify we will neglect the volume of water in the pipes.
Task
Write a program which for each data set:
reads the description of cisterns and the volume of water,
computes the level to which the cisterns will be filled with the given amount of water,
writes the result.
Input
The first line of the input contains the number of data sets k, 1 <= k <= 30. The data sets follow.
The first line of each data set contains one integer n, the number of cisterns, 1 <= n <= 50 000. Each of the following n lines consists of 4 nonnegative integers, separated by single spaces: b, h, w, d - the base level of the cistern, its height, width and depth in meters, respectively. The integers satisfy 0 <= b <= 10^6 and 1 <= h * w * d <= 40 000. The last line of the data set contains an integer V - the volume of water in cubic meters to be injected into the network. Integer V satisfies 1 <= V <= 2 * 10^9.
Output
The output should consist of exactly d lines, one line for each data set.
Line i, 1 <= i <= d, should contain the level that the water will reach, in meters, rounded up to two fractional digits, or the word 'OVERFLOW', if the volume of water exceeds the total capacity of the cisterns.
Sample Input
3
2
0 1 1 1
2 1 1 1
1
4
11 7 5 1
15 6 2 2
5 8 5 1
19 4 8 1
132
4
11 7 5 1
15 6 2 2
5 8 5 1
19 4 8 1
78 Sample Output
1.00
OVERFLOW
17.00 Source
Central Europe 2001 思路： 二分
1 #include <cstdio>
2 #include <algorithm>
3 #include <cstring>
4 #include <cmath>
5
6 using namespace std;
7
8 #define MP make_pair
9 #define PB push_back
10 #define lc (o<<1)
11 #define rc (o<<1|1)
12 typedef long long LL;
13 typedef pair<int,int> PII;
14 const double eps=1e-8;
15 const double pi=acos(-1.0);
16 const int K=5e4+7;
17 const int mod=1e9+7;
18
19 struct node
20 {
21 double b,h,s;
22 bool operator < (const node &ta) const
23 {
24 return b<ta.b;
25 }
26 }cis[K];
27 int n;
28 double v;
29 int sgn(double ta,double tb)
30 {
31 if(fabs(ta-tb)<eps) return 0;
32 return ta<tb?-1:1;
33 }
34 bool check(double x)
35 {
36 double ret=0;
37 for(int i=1;i<=n;i++)
38 {
39 if(sgn(cis[i].b,x)>0) break;
40 ret+=cis[i].s*min(cis[i].h,x-cis[i].b);
41 }
42 return sgn(ret,v)>=0;
43 }
44 int main(void)
45 {
46 int t;
47 scanf("%d",&t);
48 while(t--)
49 {
50 scanf("%d",&n);
51 double sum=0,l=1e9,r=-1e9;
52 for(int i=1;i<=n;i++)
53 {
54 double b,h,w,d;
55 scanf("%lf%lf%lf%lf",&b,&h,&w,&d);
56 cis[i]=(node){b,h,w*d};
57 sum+=h*w*d;
58 l=min(l,b),r=max(r,b+h);
59 }
60 sort(cis+1,cis+1+n);
61 scanf("%lf",&v);
62 if(sgn(v,sum)>0)
63 {
64 printf("OVERFLOW\n");
65 continue;
66 }
67 while(r-l>0.001)
68 {
69 double mid=(l+r)/2;
70 if(check(mid))
71 r=mid;
72 else
73 l=mid;
74 }
75 printf("%.2f\n",l);
76 }
77 return 0;
78 } 转载于:https://www.cnblogs.com/weeping/p/7652364.html