hdu6106--Classes

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

224 5 4 4 3 2 25 3 1 2 0 0 020 4 10 2 3 4 96 12 6 3 5 3 2

 

Sample Output

715
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.
 

 

题意

有三门课程ABC，现已知参加课程A、B、C的人数为a、b、c

同时参加AB、BC、AC的人数为d，e，f

ABC均参加的人数为g

现给出了a-g的数值，求参加课程的总人数为多少

给出的a-g中可能是假数据，求所有可能为真数据的数据中，总人数的最大值

分析

有点类似于荣斥定理的感觉，可以根据d e f和g先求出只参加AB、BC、AC的人数ab,bc,ac

如果有任何一个为负数，那么一定是假数据，不用进行下面计算了

然后根据上面的数据，分别算出只参加A、B、C的人数aa,bb,cc

同样需要非负

然后可以得到总的人数sum=aa+bb+cc+ab+ac+g

可以参考一下之前写的荣斥定理的博客：http://blog.csdn.net/Smile__wei/article/details/76506428

代码

#include<cstdio>#include<iostream>using namespace std;int main(){    int t;    scanf("%d",&t);    while(t--)    {        int ans=0;        int n;        int d[10];        scanf("%d",&n);        for(int i=0;i<n;i++)        {            for(int i=0;i<7;i++)scanf("%d",&d[i]);            int ab=d[3]-d[6],bc=d[4]-d[6],ac=d[5]-d[6];            if(ab<0||bc<0||ac<0)continue;            int a=d[0]-ab-ac-d[6];            int b=d[1]-ab-bc-d[6];            int c=d[2]-bc-ac-d[6];            if(a<0||b<0||c<0)continue;            ans=max(ans,a+b+c+ab+ac+bc+d[6]);        }        cout<<ans<<endl;    }    return 0;}

链接

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