UVALive 7267

Sackler Museum of Art and Archaeology at Peking University is located on a beautiful site near the

West Gate of Peking University campus, and its architecture takes the inspiration from buildings that

already exist on campus.

The collection of Chinese art and artifacts currently housed in this new museum contains more than

10

;

000

objects and spans a period of

280

;

000

years, from Paleolithic hominids and stone tool remains

to costumes, ceramics and paintings of the present era. The collection, which is used for teaching and

research purposes, has been acquired during the past seventy years from diverse sources.

The use of some objects in the museum remains unknown. For example, there are four pieces of

rectangular bones which can be dated back to

200

;

000

years ago, and no one knows what they were

made for. A former NOIer and present ACMer, Mr. Liang in the School of Archaeology and Museology

is very interested in those bones, and his tutor told him to use his wildest imagination to guess the

usage of them. So, one day, a crazy idea came up to him: were those bones some kind of IQ test tools

used by ancient people? Maybe the one who could pick exactly three pieces of those bones to form a

larger rectangle was considered smart at that time. So Mr. Liang wanted to write a program to find

out how to pass this IQ test imagined by him. Can you also do this?

Input

There are several test cases. The first line of input is an integer

T

(

1



T



20

), indicating the number

of test cases. Each test case is in one line and contains 8 integers describing 4 rectangular bones. Each

bone is described by 2 integers indicating its width and height.

All integers in the input are between

[1

;

1000]

.

Output

For each test case, if Mr. Liang can form a rectangle using those bones in the way he imagined, please

print ‘

Yes

’. If he can’t, print ‘

No

’ instead. Please note that the area of the new rectangle he forms must

be equal to the total area of the bones he picks.

Sample Input

2

1 1 1 1 1 2 2 2

1 1 2 2 10 10 20 20

Sample Output

Yes

No

给4个矩形；问能不能从其中任选3个形成一个大的矩形

思路:暴力

任选其中2个看能不能匹配，再去和第三个匹配

#include <bits/stdc++.h>
using namespace std;
int cw,ch;
struct rec
{
    int w,h;

};
rec a[10];
bool judge(int i,int j)
{
    for(int k=1; k<=4; k++)
    {
        if(i==k||k==j)
            continue;
        if(ch==a[k].h||cw==a[k].w||ch==a[k].w||cw==a[k].h)
            return true;
    }
    return false;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=1; i<=4; i++)
        {
            scanf("%d%d",&a[i].w,&a[i].h);
        }
        bool flag=false;
        for(int i=1; i<=4; i++)
        {
            for(int j=1; j<=4; j++)
            {
                if(i==j)
                    continue;
                if(a[i].w==a[j].w)
                {
                    cw=a[i].w;
                    ch=a[i].h+a[j].h;
                    if(judge(i,j))
                        flag=true;

                }
                if(a[i].h==a[j].w)
                {
                    cw=a[j].w;
                    ch=a[i].w+a[j].h;
                    if(judge(i,j))
                        flag=true;
                }
                if(a[i].w==a[j].h)
                {
                    cw=a[j].h;
                    ch=a[i].h+a[j].w;
                    if(judge(i,j))
                        flag=true;

                }
                if(a[i].h==a[j].h)
                {
                    ch=a[i].h;
                    cw=a[i].w+a[j].w;
                    if(judge(i,j))
                        flag=true;
                }

            }
        }
        if(flag)
            printf("Yes\n");
        else
            printf("No\n");
    }
}