计数方法，博弈论（扫描线，树形SG）：HDU 5299 Circles Game

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
　　这道题目可以说是模板题……
　　就是那个SG函数的求法貌似是结论，我不会证明。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <set>
using namespace std;
const int N=20010,M=40010;
int cnt,fir[N],to[N],nxt[N];
void addedge(int a,int b){
    nxt[++cnt]=fir[a];
    to[fir[a]=cnt]=b;
}
int sqr(int x){return x*x;}
int tmp;
struct Circle{
    int x,y,r;
    Circle(int _=0,int __=0,int ___=0){x=_;y=__;r=___;}
}c[N];

struct Point{
    int x,id,tp;
    Point(int _=0,int __=0,int ___=0){x=_;id=__;tp=___;}
    friend bool operator<(Point a,Point b){
        double x=c[a.id].y+a.tp*sqrt(sqr(c[a.id].r)-sqr(tmp-c[a.id].x));
        double y=c[b.id].y+b.tp*sqrt(sqr(c[b.id].r)-sqr(tmp-c[b.id].x));
        if(x!=y)return x<y;return a.tp<b.tp;
    } 
}st[M];

bool cmp(Point a,Point b){return a.x<b.x;}
set<Point>s;set<Point>::iterator it;

int fa[N],sg[N];
int DFS(int x){
    sg[x]=0;
    for(int i=fir[x];i;i=nxt[i])
        sg[x]^=DFS(to[i])+1;
    return sg[x];
}

void Init(){
    memset(fir,0,sizeof(fir));
    cnt=0;
}
int T,n;
int main(){
    scanf("%d",&T);
    while(T--){
        Init();scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d%d%d",&c[i].x,&c[i].y,&c[i].r);
            st[2*i-1]=Point(c[i].x-c[i].r,i,-1);
            st[2*i]=Point(c[i].x+c[i].r,i,1);
        }
        sort(st+1,st+2*n+1,cmp);
        for(int i=1;i<=2*n;i++){
            Point x=st[i];tmp=x.x;
            if(x.tp==-1){
                it=s.upper_bound(Point(0,x.id,1));
                if(it==s.end())addedge(fa[x.id]=0,x.id);
                else{
                    Point y=*it;
                    if(y.tp==1)
                        addedge(fa[x.id]=y.id,x.id);
                    else
                        addedge(fa[x.id]=fa[y.id],x.id);
                }
                s.insert(Point(0,x.id,1));
                s.insert(Point(0,x.id,-1));
            }
            else{
                s.erase(Point(0,x.id,1));
                s.erase(Point(0,x.id,-1));
            }
        }
        printf(DFS(0)?"Alice
":"Bob
");    
    }
    return 0;
}