ZOJ3370. Radio Waves(2-sat)

算是很裸的吧,二分半径,然后2-sat判定能不能满足条件。

#include <bits/stdc++.h>
using namespace std;
const int maxn=1222;
const double eps=1e-9;
struct node{
    double x,y;
}p[maxn];
struct twoset{
    int n;
    vector<int>g[maxn<<1];
    bool mark[maxn<<1];
    int s[maxn<<1],c;
    bool dfs(int x){
        if(mark[x^1])return false;
        if(mark[x])return true;
        mark[x]=true;
        s[c++]=x;
        for(int i=0;i<g[x].size();i++)if(!dfs(g[x][i]))return false;
        return true;
    }
    void init(int n){
        this->n=n;
        for(int i=0;i<n*2;i++)g[i].clear();
        memset(mark,0,sizeof mark);
    }
    void add_clause(int x,int xval,int y,int yval){
        x=x*2+xval;
        y=y*2+yval;
        g[x^1].push_back(y);
        g[y^1].push_back(x);
    }
    bool solve(){
        for(int i=0;i<n*2;i+=2){
            if(!mark[i]&&!mark[i+1]){
                c=0;
                if(!dfs(i)){
                    while(c>0)mark[s[--c]]=false;
                    if(!dfs(i+1))return false;
                }
            }
        }
        return true;
    }
}solver;

double getdis(node a,node b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void run(int n){
    for(int i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
    double l=0,r=30000,ans=0;
    while(l+eps<=r){
        double mid=(l+r)/2;
        solver.init(n);
        for(int i=0;i<n;i++)
        for(int j=i+1;j<n;j++){
            double dis=getdis(p[i],p[j]);
            if(dis<2*mid){
                solver.add_clause(i,0,j,0);
                solver.add_clause(i,1,j,1);
            }
        }
        if(solver.solve())ans=mid,l=mid+eps;
        else r=mid-eps;
    }
    solver.init(n);
    for(int i=0;i<n;i++)
    for(int j=i+1;j<n;j++){
        double dis=getdis(p[i],p[j]);
        if(dis<2*ans){
            solver.add_clause(i,0,j,0);
            solver.add_clause(i,1,j,1);
        }
    }
    solver.solve();
    printf("%.8lf
",ans);
    for(int i=0;i<n;i++){
        if(solver.mark[i<<1])putchar('1');
        else putchar('2');
        printf("%c",i==n-1?'
':' ');
    }
}
int main()
{
//    freopen("in","r",stdin);
    int n;
    while(scanf("%d",&n)>0)run(n);
    return 0;
}

  

原文地址:https://www.cnblogs.com/wshh/p/4029828.html