【2017多校】HDU 6106 Classes 【集合计算】

Classes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 698    Accepted Submission(s): 444

Problem Description

The school set up three elective courses, assuming that these courses are A, B, C. N classes of students enrolled in these courses.
Now the school wants to count the number of students who enrolled in at least one course in each class and records the maximum number of students.
Each class uploaded 7 data, the number of students enrolled in course A in the class, the number of students enrolled in course B, the number of students enrolled in course C, the number of students enrolled in course AB, the number of students enrolled in course BC, the number of students enrolled in course AC, the number of students enrolled in course ABC. The school can calculate the number of students in this class based on these 7 data.
However, due to statistical errors, some data are wrong and these data should be ignored.
Smart you must know how to write a program to find the maximum number of students.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with one integer N, indicates the number of class.
Then N lines follow, each line contains 7 data: a, b, c, d, e, f, g, indicates the number of students enrolled in A, B, C, AB, BC, AC, ABC in this class. 
It's guaranteed that at least one data is right in each test case.

Limits
T≤100
1≤N≤100
0≤a,b,c,d,e,f,g≤100

 

Output

For each test case output one line contains one integer denotes the max number of students who enrolled in at least one course among N classes.

 

Sample Input

2 2 4 5 4 4 3 2 2 5 3 1 2 0 0 0 2 0 4 10 2 3 4 9 6 12 6 3 5 3 2

 

Sample Output

7 15

Hint

In the second test case, the data uploaded by Class 1 is wrong. Because we can't find a solution which satisfies the limitation. As for Class 2, we can calculate the number of students who only enrolled in course A is 2, the number of students who only enrolled in course B is 6, and nobody enrolled in course C, the number of students who only enrolled in courses A and B is 1, the number of students who only enrolled in courses B and C is 3, the number of students who only enrolled in courses A and C is 1, the number of students who enrolled in all courses is 2, so the total number in Class 2 is 2 + 6 + 0 + 1 + 3 + 1 + 2 = 15.

 

Source

2017 Multi-University Training Contest - Team 6

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=6106

题意：n个班每个班的学生报名ABC三种课程，求n个班中人数最多的班级。

每个班级一次给出报了A,B,C,AB,AC,BC,ABC,的人数

注：数据有可能是假的，假的数据直接忽略掉，保证一定有一个真的数据

做法：1

只报AB的人数 == 报了AB的人数 -- 报了ABC的人数

只报了A的人数 == 报了A的人数 -- 只报AB的人数 -- 只报AC的人数 -- 报了ABC的人数

这样就可以求出每个类的人数。

 AC代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 

AC代码2：

刚开始有个问题，不能直接 判ab<=min(a,b) bc<=min(b,c) acwa，因为这个判断太松了，

所以应为ab+ac-abc<=a bc+ab-abc<=b ac+bc-abc<=c 

/**
  * 行有余力，则来刷题！
  * 博客链接:http://blog.csdn.net/hurmishine
  * 个人博客网站:http://wuyunfeng.cn/
*/
#include 
#include 
#include 
#include 
using namespace std;
const int maxn=1e1+5;
int a[7];
int getAns()
{
    //WA
    //if(a[6]>min(a[3],min(a[4],a[5])))
        //return 0;
    if(a[6]>min(a[3],min(a[4],a[5])))
        return 0;
    if(a[3]+a[5]-a[6]>a[0])
        return 0;
    if(a[3]+a[4]-a[6]>a[1])
        return 0;
    if(a[4]+a[5]-a[6]>a[2])
        return 0;
    return a[0]+a[1]+a[2]-a[3]-a[4]-a[5]+a[6];
}
int main()
{
    //freopen("C:\\Users\\hncu_acm\\Desktop\\data.txt","r",stdin);
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        int ans=0;
        while(n--)
        {
            for(int i=0;i<7;i++)
            {
                cin>>a[i];
            }
            ans=max(ans,getAns());
        }
        cout<

参考博客：http://blog.csdn.net/archger/article/details/77073837