HDU 5299 (树删边博弈)

Circles Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1446    Accepted Submission(s): 456

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
   
   
   
   

 

所有的圆之间构成一棵数的关系,然后就是树上的删边博弈.

建树可以扫描线,然后暴力出奇迹~

#include <bits/stdc++.h>
using namespace std;
#define maxn 41111
#define eps 1e-8

struct circle {
    double x, y, r;
    bool operator < (const circle &a) const {
        return r < a.r;
    }
}p[maxn];
int n;
int cnt = 0;
struct node {
    int u, v, next;
}edge[maxn*4];
int tot, head[maxn], fa[maxn];

bool in (int i, int j) {//判断圆i是不是在j中
    double xx = p[i].x-p[j].x;
    double yy = p[i].y-p[j].y;
    return xx*xx+yy*yy <= p[j].r*p[j].r;
}

void add_edge (int u, int v) {
    edge[tot].u = u, edge[tot].v = v, edge[tot].next = head[u], head[u] = tot++;
}

int dfs (int u) {
    int ans = 0;
    for (int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].v;
        ans ^= (dfs (v)+1);
    }
    return ans;
}

int main () {
    //freopen ("in.txt", "r", stdin);
    int t;
    scanf ("%d", &t);
    while (t--) {
        scanf ("%d", &n);
        memset (head, -1, sizeof head);
        for (int i = 0; i < n; i++) fa[i] = i;
        tot = cnt = 0;
        for (int i = 0; i < n; i++) {
            scanf ("%lf%lf%lf", &p[i].x, &p[i].y, &p[i].r);
        }
        sort (p, p+n);
        for (int i = 0; i < n; i++) {
            for (int j = i+1; j < n; j++) {
                if (in (i, j)) {
                    fa[i] = j;
                    add_edge (j, i);
                    break;
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < n; i++) if (fa[i] == i) {
            ans ^= (dfs (i)+1);
        }
        printf ("%s\n", ans ? "Alice" : "Bob");
    }
    return 0;
}