zoj 2318 Get Out! 夜

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1318

推荐解题报告:http://www.cppblog.com/Yuan/archive/2010/05/02/114163.html

自己对几何题目是一窍不通呀,尤其是精度问题 狂晕,这题也是看了别人的解析

思路,把人所在位置移动到原点,其他点也移动相应位置,然后把人的船当成点,这样的话其他岛屿的半径统一加上船的半径就可以了

然后相交的岛屿连线,看能不能有一个多边形把原点包起来

判别方法,根据点积求角度,根据叉积求正负,然后看有没有负环

更新时dist[i][j]>dist[i][l]+dist[l][j]+eps;  不是很明白为什么要加eps

代码:

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<vector>
#include<string>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<cmath>

#define LL long long

using namespace std;

const int INF=0x3f3f3f3f;
const double FINF=1e9;
const double eps=1e-6;
const double PI=acos(-1.0);
const int N=305;
struct node
{
    double x,y,r;
}circle[N];
double dist[N][N];
double Fdist(const node& a,const node& b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double crossProduct(const node& a,const node& b)
{
    return a.x*b.y-a.y*b.x;
}
double dotProduct(const node& a,const node& b)
{
    return a.x*b.x+a.y*b.y;
}
int main()
{
    double a=1.0,b=2.5,c=5.0;
    while(c>a+b)
    c=a+b;
    cout<<c<<endl;
    //freopen("data.in","r",stdin);
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        for(int i=1;i<=n;++i)
        cin>>circle[i].x>>circle[i].y>>circle[i].r;
        cin>>circle[0].x>>circle[0].y>>circle[0].r;
        for(int i=1;i<=n;++i)
        {
            circle[i].x-=circle[0].x;
            circle[i].y-=circle[0].y;
            circle[i].r+=circle[0].r;
        }
        circle[0].x=0.0;
        circle[0].y=0.0;
        circle[0].r=0.0;
        for(int i=1;i<=n;++i)
        for(int j=i;j<=n;++j)
        {
            if(i==j||circle[i].r+circle[j].r-eps<Fdist(circle[i],circle[j]))
            dist[i][j]=dist[j][i]=INF;
            else
            {
                double C=acos(dotProduct(circle[i],circle[j])/
                              (Fdist(circle[i],circle[0])*Fdist(circle[j],circle[0])));
                bool flag=(crossProduct(circle[i],circle[j])>=0.0)?true:false;
                dist[i][j]=(flag==true)?C:-C;
                dist[j][i]=-dist[i][j];
            }
        }
        for(int l=1;l<=n;++l)
        for(int i=1;i<=n;++i)
        for(int j=1;j<=n;++j)
        if(dist[i][j]>dist[i][l]+dist[l][j]+eps)
        dist[i][j]=dist[i][l]+dist[l][j]+eps;
        bool flag=true;
        for(int i=1;i<=n;++i)
        if(dist[i][i]<-2*PI)
        {flag=false;break;}
        if(flag==true)
        cout<<"YES"<<endl;
        else
        cout<<"NO"<<endl;
        if(T>0)
        cout<<endl;
    }
    return 0;
}

  

原文地址:https://www.cnblogs.com/liulangye/p/3055140.html