首页 > 代码库 > hdu5299 Circles Game
hdu5299 Circles Game
题意是这样。给出非常多圆,要么两两相离,要么包括,若删掉一个圆,那被他包括的都要删除,若某人没有圆给他删,那么他就赢了。
。。。知道树上博弈的话。就非常easy。
。。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能够包括它的圆,建立j到i的连边,然后break
这样就建立好了一棵树,之后知道这个就非常easy了。。。
树的删边游戏
规则例如以下:
? 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。
我们有例如以下定理:
[定理]
叶子节点的 SG 值为 0;
中间节点的 SG 值为它的全部子节点的 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.
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。
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
2015 Multi-University Training Contest 1
hdu5299 Circles Game
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。