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I have come here to ask help to dp experts. I will highly appreciate it if you can help solve this dilemma. I have been struggling with this for a while now. ↵
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The problem can be found here.↵
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http://mirror.codeforces.com/contest/69/problem/D↵
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I have stepped through the logic of my DP formulation and i dont seem to spot the fallacy. So, can you please point out the point where my logic fails. I will highly appreciate it. Thanks for your time. I have commented out parts of my code for legibility. ↵
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The test case that fails is the following↵
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-100 -100 10 200↵
0 1↵
1 0↵
1 1↵
31 41↵
3 4↵
5 2↵
1 2↵
3 3↵
9 8↵
10 2↵
~~~~~↵
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Code starts below↵
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~~~~~↵
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import math↵
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vectors = []↵
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line_parts = raw_input().split()↵
x = int(line_parts[0])↵
y = int(line_parts[1])↵
n = int(line_parts[2])↵
d = int(line_parts[3])↵
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for i in range(n):↵
l = raw_input().split()↵
vectors.append((int(l[0]), int(l[1])))↵
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mp = {}↵
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def canWin(x, y, t, fs, ss):↵
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if math.sqrt(math.pow(x, 2) + math.pow(y, 2)) > float(d):↵
return False↵
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if (x, y, t) in mp:↵
return mp[(x, y, t)] ↵
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result = True ↵
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if t == False and ss == False: #if Dasha turn and if dasha hasnt swapped then can swap 1 time↵
for v in vectors:↵
result = result and (not (canWin(x + v[0], y + v[1], not t, fs, ss)))↵
result = result and (not canWin(y, x, not t, fs, True)) #swap step↵
elif t == True and fs == False: #if Anton turn and if Anton hasnt swapped then can swap 1 time↵
for v in vectors:↵
result = result and (not (canWin(x + v[0], y + v[1], not t, fs, ss)))↵
result = result and (not canWin(y, x, not t, True, ss)) #swap step↵
else: #since both has swapped 1 time now cannot swap further↵
for v in vectors:↵
result = result and (not (canWin(x + v[0], y + v[1], not t, fs, ss)))↵
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mp[(x, y, t)] = result↵
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return result↵
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result = False↵
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for v in vectors:↵
result = result or canWin(x + v[0], y + v[1], False, False, False) ↵
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if result == True:↵
print "Anton"↵
else:↵
print "Dasha"↵
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~~~~~↵
