Time Limit: 1.0 Seconds Memory Limit: 65536K
Total Runs: 846 Accepted Runs: 304 Multiple test files
Joe delivers pallets for Ivan. Joe is not very smart and often makes mistakes — he brings Ivan pallets that do not fit together to make a box. But Joe does not trust Ivan. It always takes a lot of time to explain Joe that he has made a mistake.
Fortunately, Joe adores everything related to computers and sincerely believes that computers never make mistakes. Ivan has decided to use this for his own advantage. Ivan asks you to write a program that given sizes of six rectangular pallets tells whether it is possible to make a box out of them.
Input
Input consists of six lines. Each line describes one pallet and contains two integer numbers w and h (1 ≤ w, h ≤ 10 000) — width and height of the pallet in millimeters respectively.
Output
Write a single word "POSSIBLE" to the output if it is possible to make a box using six given pallets for its sides. Write a single word "IMPOSSIBLE" if it is not possible to do so.
| Sample Input | Sample Output |
|---|---|
1345 2584 2584 683 2584 1345 683 1345 683 1345 2584 683 |
POSSIBLE |
1234 4567 1234 4567 4567 4321 4322 4567 4321 1234 4321 1234 |
IMPOSSIBLE |
Source: Northeastern European 2004
题解： ：对面进行排序，比较相邻的两个面是否相同，当两个面相同的时候看第一个面和第三个面是否有公共边，最后看最后的面可以和前两个面连起来吗
重复的东西可以写成函数 不易出错，
这里注意，如果最后满足题意的有一种情况的时候，flag定义成false，有一个满足就更新成true，
而要找所有条件都满足的时候，开始将flag定义成true，有一个不满足就更新成false
其实刷水题还是有用的，gaga.
1 #include <cstdio> 2 #include <algorithm> 3 using namespace std; 4 5 struct pg{ 6 int x, y; 7 bool operator < (const pg a) const { 8 return x == a.x ? y < a.y : x < a.x; 9 } 10 }p[6]; 11 bool ch(int x, int y, pg tm) 12 { 13 if(x > y) swap(x, y); 14 return x == tm.x && y == tm.y; 15 } 16 17 int main() 18 { 19 while(~scanf("%d %d", &p[0].x, &p[0].y)) 20 { 21 for(int i = 1; i < 6; i++) scanf("%d %d", &p[i].x, &p[i].y); 22 for(int i = 0; i < 6; i++) if(p[i].x > p[i].y) swap(p[i].x, p[i].y); 23 sort(p, p+6); 24 if(p[0].x != p[1].x || p[0].y != p[1].y || p[2].x != p[3].x || p[2].y != p[3].y || p[4].x != p[5].x || p[4].y != p[5].y) 25 { 26 puts("IMPOSSIBLE"); 27 continue; 28 } 29 pg a = p[0], b = p[2], c = p[4]; 30 int t1, t2; 31 bool flag = false; 32 if(a.x == b.x && ch(a.y, b.y, c)) flag = true; 33 if(a.x == b.y && ch(a.y, b.x, c)) flag = true; 34 if(a.y == b.x && ch(a.x, b.y, c)) flag = true; 35 if(a.y == b.y && ch(a.x, b.x, c)) flag = true; 36 if(flag) puts("POSSIBLE"); 37 else puts("IMPOSSIBLE"); 38 } 39 return 0; 40 }