poj 2420/1379 随机/模拟退火（求费马点/最偏离点）

2420题意：给出平面中一些点，求平面中的一个点，使得这个点到其他所有点的距离之和最短(这个点叫作费马点)，并输出这个距离和。

1379题意：背景和2420相同，只是求一个坐标到其他所有点的最短距离最大。

思路：先随机取个点，再取一个步长，朝4个方向走。如果新位置到各点距离比原来小，就走过去；直到当前位置比四周都好，此时缩小步长。如此迭代，直到步长小于题目的精度（这道题是0.1）输出即可。实际上缩小步长蕴含了模拟退火的思想。不过模拟退火中还可以按照一定概率留在原地（即使移动可以获得更好的效益值）。

朝四个方向走可以改为朝四周所有方向走，（http://www.2cto.com/kf/201409/334851.html）采用的是这种方法。

while(L > EPS) {
        double alpha = 2 * PI * Rand();//一个随机的角度
        Point temp(now.x + cos(alpha) * T,now.y + sin(alpha) * T);
        dE = Statistic(now) - Statistic(temp);
        if(dE >= 0 || exp(dE / L) > Rand)
            now = temp;
        L *= .99;
    }

2420代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define N 105
struct point{
    double x,y;
}s[N],now,tmp;
int n;
int ori[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
double dd(struct point a,struct point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double getdis(struct point now){
    int i;
    double res=0;
    for(i = 0;i<n;i++)
        res += dd(s[i],now);
    return res;
}
int main(){
    int i;
    double j,left,right,up,down,diff,dis;
    left = down = 10000;
    right = up = 0;
    scanf("%d",&n);
    for(i = 0;i<n;i++){
        scanf("%lf %lf",&s[i].x,&s[i].y);
        left = min(left,s[i].x);
        right = max(right, s[i].x);
        up = max(up, s[i].y);
        down = min(down, s[i].y);
    }
    now.x = (left+right)/2;//随机取个点，尽量往中间取
    now.y = (up+down)/2;
    diff = max(right-left,up-down) / 2;//取一个步长
    dis = getdis(now);
    while(diff > 0.1){
        for(i = 0;i<4;i++){
            tmp.x = now.x + ori[i][0]*diff;
            tmp.y = now.y + ori[i][1]*diff;
            if((j=getdis(tmp)) < dis){//如果四周优于当前点就走过去
                now = tmp;
                dis = j;
                break;
            }
        }
        if(i==4)//减小步长
            diff *= 0.5;
    }
    printf("%.0f\n",dis);
    return 0;
}

1379代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <ctime>
#include <cstdlib>
#define clr(s,t) memset(s,t,sizeof(s))
using namespace std;
#define INF 0x3fffffff
struct point{
    double x,y;
}p[1005],res,now,tmp;
int n,m,h,T;
int ori[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};
int check(double x,double y){
    return x>=0 && y>=0 && x<=n && y<=m;
}
double dd(point a,point b){
    return (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);
}
int dist(point a){
    double res = 0x3fffffff;
    for(int i = 1;i<=h;i++)
        res = min(res,dd(a,p[i]));
    return res;
}
int cmp(point a,point b){
    if(a.x == b.x)
        return a.y < b.y;
    return a.x < b.x;
}
int eq(point a,point b){
    return a.x==b.x && a.y==b.y;
}
int main(){
    scanf("%d",&T);
    srand((unsigned int)time(0));
    while(T--){
        int i,j,k;
        double off;
        res.x = res.y = 0;
        scanf("%d %d %d",&n,&m,&h);
        for(i = 1;i<=h;i++)
            scanf("%lf %lf",&p[i].x,&p[i].y);
        sort(p+1, p+1+h, cmp);
        h = unique(p+1, p+1+h, eq)-p-1;
        for(k = 1;k<=20;k++){//多取几次初始点可以增大AC概率
            now.x = rand()%n;
            now.y = rand()%m;
            for(off = min(m,n);off>=0.001;){
                for(j = 0;j<4;j++){
                    tmp.x = now.x+ori[j][0]*off;
                    tmp.y = now.y+ori[j][1]*off;
                    if(check(tmp.x,tmp.y) && dist(tmp)>dist(now)){
                        now = tmp;
                        break;
                    }
                }
                if(j==4)
                    off /= 2;
            }
            if(dist(res) < dist(now))
                res = now;
        }
        printf("The safest point is (%.1lf, %.1lf).\n",res.x,res.y);
    }
    return 0;
}