标签:ica size sed mil iostream include sts style namespace
将全部的圆化成树,然后就能够转化成树上的删边博弈问题....
2 1 0 0 1 6 -100 0 90 -50 0 1 -20 0 1 100 0 90 47 0 1 23 0 1
Alice Bob
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <vector> using namespace std; #pragma comment(linker, "/STACK:1024000000,1024000000") const int maxn=20200; int n; struct Circle { int x,y,r; bool operator<(const Circle& cir) const { return r>cir.r; } }circle[maxn]; double dist(int a,int b) { return sqrt((circle[a].x-circle[b].x)*(circle[a].x-circle[b].x) +(circle[a].y-circle[b].y)*(circle[a].y-circle[b].y)); } vector<int> edge[maxn]; void Link(int u,int x) { bool fg=true; for(int i=0,sz=edge[u].size();i<sz;i++) { int v=edge[u][i]; double dd=dist(x,v); if(dd+circle[x].r>circle[v].r) continue; fg=false; Link(v,x); return ; } if(fg) edge[u].push_back(x); } int dp[maxn]; void dfs(int u) { int ret=-1; for(int i=0,sz=edge[u].size();i<sz;i++) { int v=edge[u][i]; dfs(v); if(ret==-1) ret=dp[v]+1; else ret^=dp[v]+1; } if(ret==-1) ret=0; dp[u]=ret; } int main() { int T_T; scanf("%d",&T_T); while(T_T--) { scanf("%d",&n); for(int i=1,x,y,r;i<=n;i++) { scanf("%d%d%d",&x,&y,&r); //circle[i]=(Circle){x,y,r}; circle[i].x=x; circle[i].y=y; circle[i].r=r; edge[i].clear(); } edge[0].clear(); sort(circle+1,circle+1+n); for(int i=1;i<=n;i++) { Link(0,i); dp[i]=0; } dfs(0); if(dp[0]==0) puts("Bob"); else puts("Alice"); } return 0; }
HDOJ 5299 Circles Game 圆嵌套+树上SG
标签:ica size sed mil iostream include sts style namespace
原文地址:http://www.cnblogs.com/liguangsunls/p/7183781.html