I am aware of the activities selection problem that can be seen here. Since we need to find for 2 people, i tried finding max possible jobs for Cameron and then removed all the jobs that can be done by Cameron. Then, in the remaining activities, I found the max possible jobs that can be done by Jamie. If there are still jobs left to be done, I say its impossible, else i print the person in order of the jobs.
该代码适用于示例情况，我自己尝试了其他几种情况，但无法找到给出错误答案的特殊情况。而且，GCJ表示它没有给出正确的答案。
在方法，思想，极端情况或完全不同的方法上的任何错误都将受到赞赏。
question can be found here and my code is
cases = int(input())
for tt in range(cases):
n = int(input())
pl1 = []
pl2 = []
activ = []
cjc = []
cjj = []
plx2 = []
plx1 = []
for xx in range(n):
p1,p2 = list(map(int,input().split()))
pl1.append(p1)
pl2.append(p2)
activ.append([p1,p2])
pl = sorted(pl2)
p02 = pl[0]
p01 = pl1[pl2.index(p02)]
cjj.append([p01,p02])
activ.remove([p01,p02])
for p2 in pl[1:]:
p1 = pl1[pl2.index(p2)]
if(p1>=p02):
cjj.append([p1,p2])
activ.remove([p1,p2])
p02=p2
if(len(activ)>0):
for i in activ:
plx1.append(i[0])
plx2.append(i[1])
pll = sorted(plx2)
p02 = pll[0]
p01 = plx1[plx2.index(p02)]
cjc.append([p01,p02])
activ.remove([p01,p02])
for p2 in pll[1:]:
p1 = plx1[plx2.index(p2)]
if(p1>=p02):
cjc.append([p1,p2])
activ.remove([p1,p2])
p02=p2
s = ''
fff = False
if(len(activ)>0):
s = "IMPOSSIBLE"
else:
s = ''
for i in range(n):
if(fff == False):
p1 = pl1[i]
p2 = pl2[i]
if([p1,p2] in cjc):
s+='C'
elif([p1,p2] in cjj):
s+='J'
else:
s = "IMPOSSIBLE"
fff = True
ss = "Case #"+str(tt+1)+":"
print(ss,s)
