HDU 2295 Radar （重复覆盖的模板 即每列至少一个1）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2295

 

题目大意：有n个城市，m个雷达，k个人，然后给出n个城市和m个雷达站的坐标

现在要求选出k个雷达站，求出能覆盖所有城市的雷达最小半径

思路：相当于n个城市为列，m个雷达为行。要求每个城市至少没被一个雷达覆盖.

二分答案， 然后使用重复覆盖的Dancing Links模板进行判断，看使用K个能不能覆盖n个点

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;const int maxnode = 3000;const int MaxM = 55;const int MaxN = 55;int K;struct DLX{    int n,m,size;    int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];    int H[MaxN],S[MaxN];    int ands,ans[MaxN];    void init(int _n,int _m)    {        n = _n;        m = _m;        for(int i = 0;i <= m;i++)        {            S[i] = 0;            U[i] = D[i] = i;            L[i] = i-1;            R[i] = i+1;        }        R[m] = 0; L[0] = m;        size = m;        for(int i = 1;i <= n;i++)            H[i] = -1;    }    void Link(int r,int c)    {        ++S[Col[++size]=c];        Row[size] = r;        D[size] = D[c];        U[D[c]] = size;        U[size] = c;        D[c] = size;        if(H[r] < 0)H[r] = L[size] = R[size] = size;        else        {            R[size] = R[H[r]];            L[R[H[r]]] = size;            L[size] = H[r];            R[H[r]] = size;        }    }    void remove(int c)    {        for(int i = D[c];i != c;i = D[i])            L[R[i]] = L[i], R[L[i]] = R[i];    }    void resume(int c)    {        for(int i = U[c];i != c;i = U[i])            L[R[i]]=R[L[i]]=i;    }    bool v[maxnode];    int f()    {        int ret = 0;        for(int c = R[0];c != 0;c = R[c])v[c] = true;        for(int c = R[0];c != 0;c = R[c])            if(v[c])            {                ret++;                v[c] = false;                for(int i = D[c];i != c;i = D[i])                    for(int j = R[i];j != i;j = R[j])                        v[Col[j]] = false;            }        return ret;    }    bool Dance(int d)    {        if(d + f() > K)return false;        if(R[0] == 0)return d <= K;        int c = R[0];        for(int i = R[0];i != 0;i = R[i])            if(S[i] < S[c])                c = i;        for(int i = D[c];i != c;i = D[i])        {            remove(i);            for(int j = R[i];j != i;j = R[j])remove(j);            if(Dance(d+1))return true;            for(int j = L[i];j != i;j = L[j])resume(j);            resume(i);        }        return false;    }};DLX g;const double eps = 1e-8;struct Point{    int x,y;    void input()    {        scanf("%d%d",&x,&y);    }}city[MaxM],station[MaxN];double dis(Point a,Point b){    return sqrt((double)(a.x-b.x)*(a.x-b.x)+(double)(a.y-b.y)*(a.y-b.y));}int main(){    int T;    int n,m;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&m,&K);        for(int i = 0;i < n;i++)city[i].input();        for(int i = 0;i < m;i++)station[i].input();        double l = 0, r = 1e8;        while(r-l >= eps)//二分查找        {            double mid = (l+r)/2;            g.init(m,n);            for(int i = 0;i < m;i++)                for(int j = 0;j < n;j++)                    if(dis(station[i],city[j]) < mid - eps)//如果小于假设的距离的话                        g.Link(i+1,j+1);            if(g.Dance(0))r = mid-eps;//如果成立的  将r缩小查询            else l = mid+eps;//如果不成立的话  将l增大查询        }        printf("%.6lf\n",l);    }    return 0;}