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  • poj 1379 Run Away 模拟退火 难度:1

    Run Away
    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 6482   Accepted: 1993

    Description

    One of the traps we will encounter in the Pyramid is located in the Large Room. A lot of small holes are drilled into the floor. They look completely harmless at the first sight. But when activated, they start to throw out very hot java, uh ... pardon, lava. Unfortunately, all known paths to the Center Room (where the Sarcophagus is) contain a trigger that activates the trap. The ACM were not able to avoid that. But they have carefully monitored the positions of all the holes. So it is important to find the place in the Large Room that has the maximal distance from all the holes. This place is the safest in the entire room and the archaeologist has to hide there.

    Input

    The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing three integers X, Y, M separated by space. The numbers satisfy conditions: 1 <= X,Y <=10000, 1 <= M <= 1000. The numbers X and Yindicate the dimensions of the Large Room which has a rectangular shape. The number M stands for the number of holes. Then exactly M lines follow, each containing two integer numbers Ui and Vi (0 <= Ui <= X, 0 <= Vi <= Y) indicating the coordinates of one hole. There may be several holes at the same position.

    Output

    Print exactly one line for each test case. The line should contain the sentence "The safest point is (P, Q)." where P and Qare the coordinates of the point in the room that has the maximum distance from the nearest hole, rounded to the nearest number with exactly one digit after the decimal point (0.05 rounds up to 0.1).

    Sample Input

    3
    1000 50 1
    10 10
    100 100 4
    10 10
    10 90
    90 10
    90 90
    3000 3000 4
    1200 85
    63 2500
    2700 2650 
    2990 100

    Sample Output

    The safest point is (1000.0, 50.0).
    The safest point is (50.0, 50.0).
    The safest point is (1433.0, 1669.8).


    思路:直接按照题解做的,首先列举30个随机点,然后分别进行短距离的随机尝试

    #include <cstdio>
    #include <ctime>
    #include <cstring>
    #include <cmath>
    #include <cstdlib>
    #include <algorithm>
    using namespace std;
    const double Pi=acos(-1.0);
    const double eps=1e-3;
    const double inf=1e20;
    double px[31],py[31],d[31],x[1111],y[1111];
    double caldis(double x1,double y1,double x2,double y2){
        return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
    }
    int main(){
        int T;
        scanf("%d",&T);
        srand(9876543);
        while(T--){
            int a,b,m;
            scanf("%d%d%d",&a,&b,&m);
            for(int i=0;i<m;i++){
                scanf("%lf%lf",x+i,y+i);
            }
            for(int i=0;i<30;i++){
                px[i]=(double)(rand()%1000+1)/1000.000*a;
                py[i]=(double)(rand()%1000+1)/1000.000*b;
                d[i]=inf;
                for(int j=0;j<m;j++)d[i]=min(d[i],caldis(px[i],py[i],x[j],y[j]));
            }
            double delta=double(max(a,b))/(sqrt(1.0*m));
            while(delta>eps){
                for(int i=0;i<30;i++){
                    double tx=px[i],ty=py[i];
                    for(int j=0;j<30;j++){
                        double theta=(double)(rand()%1000+1)/1000.000*10*Pi;
                        double dx=delta*cos(theta);
                        double dy=delta*sin(theta);
                        tx+=dx;ty+=dy;
                        if(tx<0||tx>a||ty<0||ty>b){tx-=dx;ty-=dy;continue;}
                        double td=inf;
                        for(int k=0;k<m;k++)td=min(td,caldis(tx,ty,x[k],y[k]));
                        if(td>d[i]){
                            d[i]=td;px[i]=tx;py[i]=ty;
                        }
                        tx-=dx;ty-=dy;
                    }
                }
                delta*=0.9;
            }
            double ans=0;int ind=0;
            for(int i=0;i<30;i++){
                if(d[i]>ans){
                    ind=i;
                    ans=d[i];
                }
            }
             printf("The safest point is (%.1f, %.1f).
    ",px[ind],py[ind]);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/xuesu/p/4111837.html
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