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Radar

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3 3 2
3 4
3 1
5 4
1 1
2 2
3 3

2.236068

有n个城市，m个发电站，已知他们的坐标，每个发电站覆盖半径相同，求多大半径时可以把所有的城市都覆盖，二分半径建图，dlx爆搜。

代码：

/* ***********************************************
Author :_rabbit
Created Time :2014/4/28 14:08:33
File Name :1.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
struct Point{
    double x,y;
}city[100],radar[100];
double dis[66][66];
double dist(Point a,Point b){
    double s=a.x-b.x,t=a.y-b.y;
    return s*s+t*t;
}
int K;
struct DLX{
    const static int maxn=20010;
    #define FF(i,A,s) for(int i = A[s];i != s;i = A[i])
    int L[maxn],R[maxn],U[maxn],D[maxn];
    int size,col[maxn],row[maxn],s[maxn],H[maxn];
    bool vis[70];
    int ans[maxn],cnt;
    void init(int m){
        for(int i=0;i<=m;i++){
            L[i]=i-1;R[i]=i+1;U[i]=D[i]=i;s[i]=0;
        }
        memset(H,-1,sizeof(H));
        L[0]=m;R[m]=0;size=m+1;
    }
    void link(int r,int c){
         U[size]=c;D[size]=D[c];U[D[c]]=size;D[c]=size;
         if(H[r]<0)H[r]=L[size]=R[size]=size;
         else {
             L[size]=H[r];R[size]=R[H[r]];
             L[R[H[r]]]=size;R[H[r]]=size;
         }
         s[c]++;col[size]=c;row[size]=r;size++;
     }
    void del(int c){//精确覆盖
        L[R[c]]=L[c];R[L[c]]=R[c];  
        FF(i,D,c)FF(j,R,i)U[D[j]]=U[j],D[U[j]]=D[j],--s[col[j]];  
    }  
    void add(int c){  //精确覆盖
        R[L[c]]=L[R[c]]=c;  
        FF(i,U,c)FF(j,L,i)++s[col[U[D[j]]=D[U[j]]=j]];  
    }  
    bool dfs(int k){//精确覆盖
        if(!R[0]){  
            cnt=k;return 1;  
        }  
        int c=R[0];FF(i,R,0)if(s[c]>s[i])c=i;  
        del(c);  
        FF(i,D,c){  
            FF(j,R,i)del(col[j]);  
            ans[k]=row[i];if(dfs(k+1))return true;  
            FF(j,L,i)add(col[j]);  
        }  
        add(c);  
        return 0;
    }  
    void remove(int c){//重复覆盖
        FF(i,D,c)L[R[i]]=L[i],R[L[i]]=R[i];
    }
     void resume(int c){//重复覆盖
         FF(i,U,c)L[R[i]]=R[L[i]]=i;
     }
    int A(){//估价函数
        int res=0;
        memset(vis,0,sizeof(vis));
        FF(i,R,0)if(!vis[i]){
                res++;vis[i]=1;
                FF(j,D,i)FF(k,R,j)vis[col[k]]=1;
            }
        return res;
    }
    void dfs(int now,int &ans){//重复覆盖
        if(R[0]==0)ans=min(ans,now);
        else if(now+A()<ans){
            int temp=INF,c;
            FF(i,R,0)if(temp>s[i])temp=s[i],c=i;
            FF(i,D,c){
                remove(i);FF(j,R,i)remove(j);
                dfs(now+1,ans);
                FF(j,L,i)resume(j);resume(i);
            }
        }
    }
/*    int astar(){
        int lim=A();
        while(!dfs(0,lim))lim++;
        return lim;
    }*/
}dlx;
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     int T,n,m;
     cin>>T;
     while(T--){
         scanf("%d%d%d",&n,&m,&K);
         for(int i=1;i<=n;i++)scanf("%lf%lf",&city[i].x,&city[i].y);
         for(int i=1;i<=m;i++)scanf("%lf%lf",&radar[i].x,&radar[i].y);
         for(int i=1;i<=m;i++)
             for(int j=1;j<=n;j++)dis[i][j]=dist(radar[i],city[j]);
         double left=0,right=1500;
         while(right-left>1e-7){
             double mid=(left+right)/2;
             double ss=mid*mid;
             dlx.init(n);
             for(int i=1;i<=m;i++)
                 for(int j=1;j<=n;j++)
                     if(dis[i][j]<=ss)dlx.link(i,j);
        //     cout<<left<<" "<<right<<" "<<ans<<" "<<dlx.size<<endl;
             int ans=INF;
             dlx.dfs(0,ans);
             if(ans>K)left=mid;
             else right=mid;
         }
         printf("%.6lf\n",left);
     }
     return 0;
}

HDU 2295 DLX 二分,码迷,mamicode.com

HDU 2295 DLX 二分

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原文地址：http://blog.csdn.net/xianxingwuguan1/article/details/24673451