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  • P1783 海滩防御

    jisoo

    简单的二分

    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cmath>
    #include<cstring>
    using namespace std;
    int n,m;
    struct po{
    	int x;
    	int y;
    }p[1005];
    int x,y;
    int r[1005];
    bool cmp(po x,po j){
    	return x.x<j.x;
    }
    double l,rr,mid;
    double ans;
    double dis[1005][1005];
    int vis[1005];
    bool dfs(int x,double mid){
    	//cout<<x<<endl;
    	if((n-p[x].x)*2<=mid){
    		return 1;
    	}else{
    		for(int i=1;i<=m;++i){
    			if(dis[x][i]<=mid&&i!=x&&!vis[i]){
    				vis[i]=1;
    				if(dfs(i,mid)) return 1;
    			}
    		}
    		return 0;
    	}
    	return 0;
    }
    bool check(double aim){
    	if(aim>=p[1].x&&aim>=(n-p[m].x)){
    		for(int i=1;i<=m;++i){
    			if(p[i].x>aim)
    			break;
    			if(vis[i]) continue;
    			vis[i]=1;
    			if(dfs(i,aim*2))
    				return 1;
    		}
    	}
    	return 0;
    }
    int main(){
    	scanf("%d%d",&n,&m);
    	for(int i=1;i<=m;++i){
    		scanf("%d%d",&p[i].x,&p[i].y); 
    	}
    	rr=n;
    //	for(int i=1;i<=m;++i){
    //		cout<<r[i]<<endl;
    //	}
    	sort(p+1,p+m+1,cmp);
    	for(int i=1;i<=m;++i){
    		for(int j=i+1;j<=m;++j){
    			double di=(double)sqrt((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y));
    			dis[i][j]=dis[j][i]=di;
    		}
    	}
    	while(l+1e-6<=rr){
    		mid=(l+rr)/2; 
    		if(check(mid)){
    			rr=mid;
    			ans=mid;
    		} else{
    			l=mid;
    		}
    		memset(vis,0,sizeof(vis));
    	}
    	printf("%.2lf",ans);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/For-Miku/p/15310286.html
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