转化为树的删边游戏。。。
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。
Circles Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 422 Accepted Submission(s): 101
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
暴力程序:
#include <bits/stdc++.h> using namespace std; #define prt(k) cerr<<#k" = "<<k<<endl typedef unsigned long long ll; const int N = 20003; struct P { int x, y, r; bool operator<(P b) const { return x < b.x; } ll d(P b) { ll dx = x - b.x, dy = y - b.y; return dx*dx + dy*dy; } }p[N]; bool vis[N]; bool cmp(P a, P b) { return a.r > b.r; } int n; inline ll sqr(ll x) { return x*x; } struct Edge { int v, next; } e[N<<1] ; int head[N], mm; void add(int u, int v) { // printf("%d --> %d ", u, v); e[mm].v = v; e[mm].next = head[u]; head[u] = mm++; } int dfs(int u) { int ret= 0; for (int i=head[u];~i;i=e[i].next) { ret ^= dfs(e[i].v)+1; } return ret; } int main() { int re, ca=1; cin>>re; while ( re-- ) { memset(head, -1, sizeof head); mm= 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=n-1;i>0;i--) { bool ok = false; for (int j=i-1;j>=0;j--) { /// r[j] > r[i] ll a = p[i].d(p[j]); ll b = sqr(p[j].r - p[i].r); if (a < b) { ok = true; add(j, i); break; } } if (!ok) { add(rt, i); } } add(rt, 0); int ans = dfs(rt); puts(ans ?"Alice" : "Bob"); } return 0; }
#include <bits/stdc++.h> using namespace std; #define prt(k) cerr<<#k" = "<<k<<endl typedef long long ll; const int N = 20003; double sqr(double x) { return x*x ; } struct P { int x, y, r; int id; P() {} P(int _x, int _y, int _r =0) { x=_x, y=_y, r=_r; } void out() { printf("%d : (%d, %d) ---%d ", id, x,y,r); } bool operator < (P b) const { if (x - b.x) return x < b.x; if (y - b.y) return y < b.y; return r < b.r; } double dis(P b) { return sqrt(sqr(x-b.x) + sqr(y-b.y) ); } }p[N]; double dis(P a, P b) { return a.dis(b); } bool cmp(P a, P b) { return a.r > b.r; } bool vis[N]; struct Edge { int v, next; } e[N<<1] ; int head[N], mm; void add(int u, int v) { e[mm].v = v; e[mm].next = head[u]; head[u] = mm++; } int n; const int Root = N - 1; int fa[N]; int dfs(int u) { int ret= 0; for (int i=head[u];~i;i=e[i].next) { ret ^= dfs(e[i].v)+1; } return ret; } set<P>::iterator it; const int inf = 0x3f3f3f3f; int main() { int re, ca=1; cin>>re; while (re--) { memset(head, -1, sizeof head); mm= 0; cin>>n; set<P> se; for (int i=0;i<n;i++) { scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].r); } se.insert(P(inf, inf)); sort(p, p+n, cmp); for (int i=0;i<n;i++) p[i].id= i ; memset(vis, 0, sizeof vis); for (int i=n-1;i>=0;i--) { P l = P(p[i].x - p[i].r, p[i].y - p[i].r); P r = P(p[i].x + p[i].r, p[i].y + p[i].r); set<P>::iterator e,f ; e = se.lower_bound(l); f = se.lower_bound(r); for (set<P>::iterator it = e; it!=se.end() && it!=f; it++) { P tmp = *it; double d = dis(tmp, p[i]); if (!vis[it->id] && d < abs(p[i].r - tmp.r)) { add(i , tmp.id); vis[tmp.id] = 1; } } se.insert(p[i]); } for (int i=0;i<n;i++) if (!vis[i]) add(Root, i); int ans = dfs(Root); puts(ans ? "Alice" : "Bob"); } }