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hdu5299 Circles Game

时间:2015-07-23 09:30:02      阅读:110      评论:0      收藏:0      [点我收藏+]

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题意是这样,给出很多圆,要么两两相离,要么包含,若删掉一个圆,那被他包含的都要删除,若某人没有圆给他删,那么他就赢了。


。。。知道树上博弈的话,就很简单。。。不知道的话,这确实是个神题……


按半径上升排序,从左往右扫,i扫到第一个j可以包含它的圆,建立j到i的连边,然后break


这样就建立好了一棵树,之后知道这个就很简单了。。。

树的删边游戏
规则如下:
 给出一个有 N 个点的树,有一个点作为树的根节点。
 游戏者轮流从树中删去边,删去一条边后,不与根节点相连的
部分将被移走。
 谁无路可走谁输。
我们有如下定理:
[定理]
叶子节点的 SG 值为 0;

中间节点的 SG 值为它的所有子节点的 SG 值加 1 后的异或和。

当然啦,像我这样暴力的写法,交c++会超时

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
struct Point
{
	int x,y,r;
}point[20010];
bool cmp(Point a,Point b)
{
	return a.r<b.r;
}
struct Edge
{
	int to,next;
}edge[20010];
int head[20010],tail;
void add(int from,int to)
{
	edge[tail].to=to;
	edge[tail].next=head[from];
	head[from]=tail++;
}
int dfs(int from)
{
	int ans=0;
	for(int i=head[from];i!=-1;i=edge[i].next)
		ans^=dfs(edge[i].to)+1;
	return ans;
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		int n;
		scanf("%d",&n);
		for(int i=0;i<n;i++)
			scanf("%d%d%d",&point[i].x,&point[i].y,&point[i].r);
		sort(point,point+n,cmp);
		tail=0;
		memset(head,-1,sizeof(head));
		for(int i=0;i<n;i++)
		{
			bool flag=0;
			for(int j=i+1;j<n;j++)
				if(ll(point[j].r*point[j].r)>ll(point[i].x-point[j].x)*(point[i].x-point[j].x)+(point[i].y-point[j].y)*(point[i].y-point[j].y))
				{
					flag=1;
					add(j,i);
					break;
				}
			if(!flag)
				add(n,i);
		}
		if(dfs(n)!=0)
			puts("Alice");
		else
			puts("Bob");
	}
}

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 593    Accepted Submission(s): 164


Problem Description
There are n circles on a infinitely large table.With every two circle, either one contains another or isolates from the other.They are never crossed nor tangent.
Alice and Bob are playing a game concerning these circles.They take turn to play,Alice goes first:
1、Pick out a certain circle A,then delete A and every circle that is inside of A.
2、Failling to find a deletable circle within one round will lost the game.
Now,Alice and Bob are both smart guys,who will win the game,output the winner‘s name.
 

Input
The first line include a positive integer T<=20,indicating the total group number of the statistic.
As for the following T groups of statistic,the first line of every group must include a positive integer n to define the number of the circles.
And the following lines,each line consists of 3 integers x,y and r,stating the coordinate of the circle center and radius of the circle respectively.
n≤20000,|x|≤20000,|y|≤20000,r≤20000。
 

Output
If Alice won,output “Alice”,else output “Bob”
 

Sample Input
2 1 0 0 1 6 -100 0 90 -50 0 1 -20 0 1 100 0 90 47 0 1 23 0 1
 

Sample Output
Alice Bob
 

Author
FZUACM
 

Source
 

版权声明:本文为博主原创文章,未经博主允许不得转载。

hdu5299 Circles Game

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原文地址:http://blog.csdn.net/stl112514/article/details/47016463

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