HDU3832 Earth Hour 2011 Multi-University Training Contest 1 - Host by HNU
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题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=3832
可以想象最优连通方案不是一条链就是一个三叉。因为是无权的,最少点数实际上就是最短路。当分叉点在特定点上的时候,三叉就退化为一条链,所以我们只需要枚举分叉点即可。
先3次bfs求出各个点到0,1,2三个区域的最短路,然后枚举点求联通最小量
#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<queue>#include<vector>#include<stack>using namespace std;#define eps 1e-6#define INF 1000000struct Point{ double x,y,r;}p[230];int n,d[3][230],map[230][230];int vis[230];double dis(Point a,Point b){ return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}void init(){ scanf("%d",&n); memset(map,0,sizeof(map)); for(int i=0;i<3;i++) for(int j=0;j<n;j++) d[i][j]=INF; d[0][0]=0; d[1][1]=0; d[2][2]=0; for(int i=0;i<n;i++) { scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].r); } for(int i=0;i<n;i++) for(int j=i+1;j<n;j++) { if(dis(p[i],p[j])-p[i].r-p[j].r<=eps) { map[i][j]=1; map[j][i]=1; // 在有公共区域的圆之间建边 } }}void bfs(int k){ memset(vis,0,sizeof(vis)); queue<int> q; q.push(k); vis[k]=1; d[k][k]=0; while(!q.empty()) { int x=q.front();q.pop(); for(int i=0;i<n;i++) if(!vis[i]&&map[x][i]) { vis[i]=1; d[k][i]=d[k][x]+1; q.push(i); } }}int main(){ int Case; scanf("%d",&Case); while(Case--) { init(); int ans=INF; bfs(0); bfs(1); bfs(2); for(int i=0;i<n;i++) //枚举分叉点(也包括链的情况) { if(d[1][i]+d[2][i]+d[0][i]+1<ans) ans=d[1][i]+d[2][i]+d[0][i]+1; } if(ans==INF)printf("-1\n"); else printf("%d\n",n-ans); }return 0;}
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