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