05 2013 档案
TopCoder SRM 176 Deranged
摘要:代码:#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<set>#include<map>#include<cmath>#include<algorithm>#include<vector>#include<cmath>#include<queue>#include<stack>//#define ull unsigned long long#defi
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TopCoder SRM 390 SetOfPatterns
摘要:自己憋了N久也没思路,看了一下别人的代码,神呀,这就是差距呀,继续努力。代码:#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<set>#include<map>#include<cmath>#include<algorithm>#include<vector>#include<cmath>#include<queue>#include<stack>/
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hdu 3870 Catch the Theves
摘要:http://acm.hdu.edu.cn/showproblem.php?pid=3870最大流 转换成 最小割 最小割 转换成 最短路代码:#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<set>#include<map>#include<cmath>#include<algorithm>#include<vector>#include<cmath>#include&l
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TopCoder SRM 382 CharmingTicketsEasy
摘要:这题的主要算法是DP 只在最后结果上用了一下容斥原理代码:#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<set>#include<map>#include<cmath>#include<algorithm>#include<vector>#include<cmath>#include<queue>#include<stack>//#define u
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SPOJ 6285. Another Game With Numbers
摘要:http://www.spoj.com/problems/NGM2/容斥原理 第四题 水! 难题各种各样,水题都一个样,无语。代码:#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<set>#include<map>#include<cmath>#include<algorithm>#include<vector>#include<cmath>#include<queu
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TopCoder TCHS 16 Divisibility
摘要:容斥原理 第三题 要注意求最小公倍数时会超 long long 的问题代码:#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<set>#include<map>#include<cmath>#include<algorithm>#include<vector>#include<cmath>#include<queue>#include<stack>//
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TopCoder SRM 477 CarelessSecretary
摘要:组合+错排代码:#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<set>#include<map>#include<cmath>#include<algorithm>#include<vector>#include<cmath>#include<queue>#include<stack>//#define ull unsigned long long
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UVa 11806 - Cheerleaders
摘要:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=226&problem=2906&mosmsg=Submission+received+with+ID+11832924容斥原理 第二题 水!代码:#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<set>
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UVa 10325 - The Lottery
摘要:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=15&problem=1266&mosmsg=Submission+received+with+ID+11830997容斥原理 第一题 水!代码:#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<set>#
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hdu 3666 THE MATRIX PROBLEM
摘要:http://acm.hdu.edu.cn/showproblem.php?pid=3666差分约束,关键在于找不等式,然后建图这个题 spfa 判断负环时,如果以更新次数大于等于N判断有负环,会超时只要判断更新次数大于sqrt(N)就可以,原理不知道代码:#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<set>#include<map>#include<cmath>#include<algorithm
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D. Olya and Graph
摘要:http://codeforces.com/contest/305/problem/D虽然给的是一颗树,其实跟树没有任何关系,仔细想一下就会发现是个数学题还是要缕清楚思路 然后再写代码:#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<cmath>#include<vector>#define ll long longusing namespace std;const in
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hdu 3634 City Planning
摘要:http://acm.hdu.edu.cn/showproblem.php?pid=3634最开始想复杂了 后来静下心来一想 原来很简单 主要还是要缕清思路代码:#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<cmath>#define ll long longusing namespace std;const int N=105;const ll MOD = 1000000007;
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hdu 3632 A Captivating Match
摘要:http://acm.hdu.edu.cn/showproblem.php?pid=3632还是做题爽kill[l][r] 表示通过一定策略 l 是否可以把 l -- r 之间的其他所以人都淘汰掉代码:#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<cmath>#define ll long longusing namespace std;const int N=105;//cons
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zoj 2319 Beautiful People
摘要:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1319这题给人的第一感觉就是最长上升子序列,按S排序,对B进行求解最长上升子序列,但是N太大o(n^2)肯定不行,所以要用优化,我是用了线段数进行优化。不是经常用类,这次用了一下,还是有很多小问题的,如果在函数内部静态申请一个局部变量对象由于对象内有很大的数组,这样就相当于在函数所占用的栈区内申请了很大数组,没有语法错误,但c++是不允许它执行的换成从堆区申请就好了,不过要记得delete否则会超内存DP思想+线段树优化代码:#include<iostream>#
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zoj 2318 Get Out!
摘要:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1318推荐解题报告:http://www.cppblog.com/Yuan/archive/2010/05/02/114163.html自己对几何题目是一窍不通呀,尤其是精度问题 狂晕,这题也是看了别人的解析思路,把人所在位置移动到原点,其他点也移动相应位置,然后把人的船当成点,这样的话其他岛屿的半径统一加上船的半径就可以了然后相交的岛屿连线,看能不能有一个多边形把原点包起来判别方法,根据点积求角度,根据叉积求正负,然后看有没有负环更新时dist[i][j]>dist[
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