2015 多校联赛 ——HDU5299(树删边)
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
给出很多个园,要么相交,要么相离。如果删掉一个圆,则会删掉其中所有的。可以转化成树的删边
(咳咳!表示并不会这个方法,学习下 而且HDU上C++会超时)
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
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> using namespace std; typedef unsigned long long ll; const int N = 20003; struct P { int x, y, r; } p[N]; bool cmp(P a, P b) { return a.r < b.r; } int n; struct Edge { int v, next; } e[N] ; int head[N], m; void add(int from ,int to) { e[m].v = to; e[m].next = head[from]; head[from] = m++; } int dfs(int u) { int temp = 0; for(int i = head[u];~i;i = e[i].next) temp ^= dfs(e[i].v) + 1; return temp; } int main() { int re; cin>>re; while ( re-- ) { memset(head, -1, sizeof head); m= 0; scanf("%d", &n); for (int i=0; i<n; i++) scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].r); sort(p ,p+n, cmp); int rt = n; for (int i=0; i<n; i++) { int flag = 0; for (int j=i+1; j<n; j++) // r[j] > r[i] { ll a = ll((p[i].x-p[j].x)*(p[i].x-p[j].x))+ll((p[i].y-p[j].y)*(p[i].y-p[j].y)); ll b = ll(p[j].r - p[i].r)*ll(p[j].r - p[i].r); if (a < b) { flag = 1; add(j, i); break; } } if (!flag) { add(rt, i); //如果找不到一个圆可以包含,则用n包含,即虚拟最大的圆,可充当根节点 } } int ans = dfs(rt); puts(ans ? "Alice" : "Bob"); } return 0; }