google code jam exercise——Rope Intranet

Round 1C 2010,C轮的测试题了。本以为难度会加大,不过第一题还是比较简单的。

Rope Intranet,统计绳子的交点个数。算法很直接,遍历所有的绳子,累计它与剩下的绳子的交点的个数。

代码如下:

#!/usr/bin/python
#encoding:UTF-8
#Filename:RopeIntranet.py

import sys

def compareWire(a,b):
    if (a[0]<b[0] and a[1]>b[1]) or (a[0]>b[0] and a[1]<b[1]):
        return 1
    else:
        return 0

inname = "input.txt"
outname = "output.txt"
if len(sys.argv)>1:
    inname = sys.argv[1]
    outname = inname.rstrip(".in")
    outname = outname + ".out"
fin = open(inname,"r")
fout = open(outname,"w")

testCaseNum = int(fin.readline().rstrip("\n"))

for caseNum in range(testCaseNum):
    wireNum = int(fin.readline().rstrip("\n"))
    wireData = []
    for wireCnt in range(wireNum):
        wire = [int(val) for val in fin.readline().rstrip("\n").split()]
        wireData.append(wire)
    crossCnt = 0
    for i in range(wireNum):
        for j in range(i+1,wireNum):
            crossCnt = crossCnt + compareWire(wireData[i],wireData[j])
    answer = "Case #%d: %d\n" %(caseNum+1,crossCnt)
    fout.write(answer)

fin.close()
fout.close()

测试通过。不过在测试large case的时候时间有点长,有更快的算法么?

 

posted @ 2012-04-09 09:51  Frandy.CH  阅读(238)  评论(0编辑  收藏  举报