通信基站(dfs回溯，思维)

Description

Input

Output

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" alt="" />

Sample Input

2 2 1 1 0 0 4 4 3 100 1 0 0 1 1 500 500

Sample Output

2.00 201.41

题解：用容斥的方法，选出所有情况，对于每种情况把选好的点和未选好的点分别放在两个集合中，对于未匹配的点搜索找区间覆盖这个点的最大距离，回溯当前距离；在结果中找最小值；

AC代码:

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int INF=0xfffffff;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
typedef long long LL;
const int MAXN=;
double lx[MAXN],ly[MAXN];
double mp[MAXN][MAXN];
int N,ta,tb;
int a[MAXN],b[MAXN];
double ans,sum;
double C,R;
double d[MAXN];
double getl(int i,int j){
    double y=ly[j]-ly[i],x=lx[j]-lx[i];
    return sqrt(x*x+y*y);
}
void solve(int m){
    if(m==){
        double res=;
        for(int i=;i<N;i++)res+=d[i];
        sum=min(sum,res);
        return;
    }
    for(int i=;i<tb;i++){
        double x=getl(a[m-],b[i]);
        double val=d[i];
        d[i]=max(d[i],x);//
        solve(m-);
        d[i]=val;//
    }
}
void work(){
    ans=INF;
    for(int i=;i<(<<N);i++){
        double temp=;
        ta=tb=;
        mem(d,);
        for(int j=;j<N;j++){
            if(i&(<<j)){
                temp+=C;
                b[tb++]=j;
            //    printf("%d ",j);
            }
            else a[ta++]=j;
        }
        //puts("");
        //for(int k=0;k<N;k++)printf("d[%d]=%.1lf ",k,d[k]);puts("");
        sum=INF;
        solve(ta);
        ans=min(ans,temp+R*sum);
    }
    printf("%.2f\n",ans);
}
int main(){
    int T;
    SI(T);
    while(T--){
        scanf("%d%lf%lf",&N,&C,&R);
        for(int i=;i<N;i++){
            scanf("%lf%lf",&lx[i],&ly[i]);
        }
        for(int i=;i<N;i++)
            for(int j=i+;j<N;j++)
                mp[i][j]=mp[j][i]=getl(i,j);
            work();
    }
    return ;
}

刚开始没考虑太多，之所以wa，因为我只是对每个点找到已经选的点的最小距离，由于这是雷达，已选的点可以覆盖多个未选的点，那么距离就是最大的那个距离，而我的可能会重复；

WA代码：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int INF=0xfffffff;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
typedef long long LL;
const int MAXN=;
double lx[MAXN],ly[MAXN];
double mp[MAXN][MAXN];
int N;
double ans;
double C,R;
double d[MAXN];
double getl(int i,int j){
    double y=ly[j]-ly[i],x=lx[j]-lx[i];
    return sqrt(x*x+y*y);
}
void work(){
    ans=INF;
    for(int i=;i<(<<N);i++){
        double temp=;
        for(int k=;k<N;k++)d[k]=INF;
        for(int j=;j<N;j++){
            if(i&(<<j)){
                temp+=C;
                for(int k=;k<N;k++){
                    d[k]=min(d[k],R*mp[k][j]);
                }
                d[j]=;
            //    printf("%d ",j);
            }
        }
        //puts("");
        for(int i=;i<N;i++)temp+=d[i];
        //for(int k=0;k<N;k++)printf("d[%d]=%.1lf ",k,d[k]);puts("");
        ans=min(ans,temp);
    }
    printf("%.2f\n",ans);
}
int main(){
    int T;
    SI(T);
    while(T--){
        scanf("%d%lf%lf",&N,&C,&R);
        for(int i=;i<N;i++){
            scanf("%lf%lf",&lx[i],&ly[i]);
        }
        for(int i=;i<N;i++)
            for(int j=i+;j<N;j++)
                mp[i][j]=mp[j][i]=getl(i,j);
            work();
    }
    return ;
}