hdu 5299(树上删边游戏) Circles Game

题意：

给你一些圆圈的圆心坐标和半径,保证这些圆圈是包含或者相离的。现在两个人做博弈操作,拿掉一个圆圈,然后这个圆圈所包含的都要移除掉。两个人一直拿下去,直到某个人找不到一个可以移除的圈他就输了。

思路

学习的姿势。这是一个树上删边游戏模型；
首先按照半径升序排序,然后对于每一个圆找第一个包含它的圆,然后连一条边。建树完成。
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;
这里可以建立成森林或者直接一棵树,没有太大区别。建成一棵树估计要方便一些吧。
可以看看爱神的这篇blog;link
以及贾志豪神牛的paper;link

参考code:

/* #pragma warning (disable: 4786) #pragma comment (linker, "/STACK:0x800000") */#include <cassert>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <sstream>#include <iomanip>#include <string>#include <vector>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <algorithm>#include <iterator>#include <utility>using namespace std;template< class T > T _abs(T n) { return (n < 0 ? -n : n); }template< class T > T _max(T a, T b) { return (!(a < b) ? a : b); }template< class T > T _min(T a, T b) { return (a < b ? a : b); }template< class T > T sq(T x) { return x * x; }template< class T > T gcd(T a, T b) { return (b != 0 ? gcd<T>(b, a%b) : a); }template< class T > T lcm(T a, T b) { return (a / gcd<T>(a, b) * b); }template< class T > bool inside(T a, T b, T c) { return a<=b && b<=c; }#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MAX(a, b) ((a) > (b) ? (a) : (b))#define F(i, n) for(int (i)=0;(i)<(n);++(i))#define rep(i, s, t) for(int (i)=(s);(i)<=(t);++(i))#define urep(i, s, t) for(int (i)=(s);(i)>=(t);--(i))#define repok(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))#define MEM0(addr) memset((addr), 0, sizeof((addr)))#define MP(x, y) make_pair(x, y)#define REV(s, e) reverse(s, e)#define SET(p) memset(pair, -1, sizeof(p))#define CLR(p) memset(p, 0, sizeof(p))#define MEM(p, v) memset(p, v, sizeof(p))#define CPY(d, s) memcpy(d, s, sizeof(s))#define READ(f) freopen(f, "r", stdin)#define WRITE(f) freopen(f, "w", stdout)#define SZ(c) (int)c.size()#define PB(x) push_back(x)#define ff first#define ss second#define ll long long#define ld long double#define pii pair< int, int >#define psi pair< string, int >#define ls u << 1#define rs u << 1 | 1#define lson l, mid, u << 1#define rson mid, r, u << 1 | 1const int INF = 0x3f3f3f3f;const double eps = 1e-9;const int MAXN = 20010;const int mod = 1000000007;struct Point{    int x,y,r;}point[MAXN];int cmp(Point a,Point b){    return a.r < b.r;}int dist(int i,int j){    return (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);}struct Edge{    int to,next;}edge[MAXN];int head[MAXN],tail;void add(int from,int to){    edge[tail].to = to;    edge[tail].next = head[from];    head[from] = tail++;}int dfs(int from){    int res = 0;    for(int i = head[from];i != -1; i = edge[i].next)        res ^= dfs(edge[i].to)+1;    return res;}int main(){    //READ("in.txt");    int t;    scanf("%d",&t);    while(t--){        int n;        scanf("%d",&n);        rep(i,0,n-1){            scanf("%d%d%d",&point[i].x,&point[i].y,&point[i].r);        }        sort(point,point+n,cmp);        tail = 0;        MEM(head,-1);        rep(i,0,n-1){            int flag = 0;            rep(j,i+1,n-1){                int rr = (ll)point[j].r * point[j].r;                int dis = dist(i,j);                if(rr > dis){                      flag = 1;                    add(j,i);                    break;                }            }            if(!flag) add(n,i);        }        if(dfs(n) != 0)            puts("Alice");        else            puts("Bob");    }    return 0;}