XM Reserves

/*hdu 5885 FFT卷积problem:给出一个n×m的格子,距离它小于等于r的那些格子都会贡献p(i,j)/(d+1), d为两个格子的距离. p[i][j]为格子的价值选择一个格子, 使得贡献和最大.solve:参考：https://async.icpc-camp.org/d/558-2016-icpc因为小于r的格子都会给当前格子贡献, 所以可以考虑先计算出所有格子对<r的其它格子的贡献.那么最后一次遍历就能找出出最大值.那么会处理出来一个 (n+r+r,m+r+r)的矩阵.然后就是如何计算, 格子(i,j)按照(x,y)移动后会到格子(i+x,j+y).坐标相加如果把所有方向处理成一个一维矩阵,那么当前点与它们相乘后的位置就是当前点能到达的目标位置(卷积，坐标相加)都转换成一维.0*M+0 0*M+1 0*M+20*M+0 0*M+1 0*M+2 1*M+0 ...   (方向)---->0*M+0 0*M+1 0*M+2 1*M+0      0*M+1 0*M+2 1*M+0求卷积后坐标之间是存在一定关系的.类似于A,B集合中有多少a+b=c那种题的思想.  (新技能get)(看代码想了很久才懂 TAT.. )用a[i*M + j]保存p(i,j), b[i*M+j]保存 1/d.c = a*b, 那么第i*M + j就是i,j的贡献和.本来最开始用的m但是发现有问题,因为一行处理后有m+2*r个点,换成M就好了....hhh-2016-09-21 21:10:30*/#pragma comment(linker,"/STACK:124000000,124000000")#include <algorithm>#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <vector>#include <map>#include <math.h>#define lson  i<<1#define rson  i<<1|1#define ll long long#define clr(a,b) memset(a,b,sizeof(a))#define key_val ch[ch[root][1]][0]using namespace std;const int maxn = 1e6 + 1000;const int inf = 0x3f3f3f3f;const ll mod = 1e9 + 7;const double eps = 1e-7;template<class T> void read(T&num){    char CH;    bool F=false;    for(CH=getchar(); CH<‘0‘||CH>‘9‘; F= CH==‘-‘,CH=getchar());    for(num=0; CH>=‘0‘&&CH<=‘9‘; num=num*10+CH-‘0‘,CH=getchar());    F && (num=-num);}int stk[70], tp;template<class T> inline void print(T p){    if(!p)    {        puts("0");        return;    }    while(p) stk[++ tp] = p%10, p/=10;    while(tp) putchar(stk[tp--] + ‘0‘);    putchar(‘\n‘);}const double PI = acos(-1.0);struct Complex{    double x,y;    Complex(double _x = 0.0,double _y = 0.0)    {        x = _x;        y = _y;    }    Complex operator-(const Complex &b)const    {        return Complex(x-b.x,y-b.y);    }    Complex operator+(const Complex &b)const    {        return Complex(x+b.x,y+b.y);    }    Complex operator*(const Complex &b)const    {        return Complex(x*b.x-y*b.y,x*b.y+y*b.x);    }};void change(Complex y[],int len){    int i,j,k;    for(i = 1,j = len/2; i < len-1; i++)    {        if(i < j) swap(y[i],y[j]);        k = len/2;        while(j >= k)        {            j-=k;            k/=2;        }        if(j < k) j+=k;    }}void fft(Complex y[],int len,int on){    change(y,len);    for(int h = 2; h <= len; h <<= 1)    {        Complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));        for(int j = 0; j < len; j+=h)        {            Complex w(1,0);            for(int k = j; k < j+h/2; k++)            {                Complex u = y[k];                Complex t = w*y[k+h/2];                y[k] = u+ t;                y[k+h/2] = u-t;                w = w*wn;            }        }    }    if(on == -1)    {        for(int i = 0; i < len; i++)            y[i].x /= len;    }}double dis(int a,int b){    return sqrt(a*a + b*b);}Complex a[2*maxn];Complex b[2*maxn];int main(){    int k;    int n,m,limit;    double x,r;    while(scanf("%d%d%lf",&n,&m,&r) != EOF)    {        limit =0, k = 1;        while(r >= limit)        {            limit ++ ;        }        limit -= 1;        int M = m +limit + limit;        while(k < (n+r+r)*(m+r+r))           k <<= 1;        for(int i = 0;i < k;i++)        {            a[i] = b[i] = 0;        }        for(int i = 0;i < n;i++)        {            for(int j = 0;j < m;j++)            {                read(x);                a[i*M + j] = x;            }        }        fft(a,k,1);        for(int i = -limit;i <= limit;i++)        {            for(int j = -limit;j <= limit;j++)            {                double len = dis(i,j);                if( r > len)                b[(i+limit) * M + j + limit] = 1.0/(len+1);            }        }        fft(b,k,1);        for(int i = 0;i < k;i++)            a[i] = a[i] * b[i];        fft(a,k,-1);        double ans = 0;        for(int i = 0;i < n;i++)        {            for(int j = 0;j < m;j++)            {                ans = max(ans,a[(i + limit) * M + j + limit].x);            }        }        printf("%.3f\n",ans);    }    return 0;}