hdu5885

来源：互联网发布：c语言结构体赋值方法编辑：程序博客网时间：2024/06/14 05:14

题意：

给出一个n*m的网格，每个格点有一个权值p(i,j)，每个格点可以对周围距离严格小于r的格点贡献权值除以欧几里得距离+1，问最后所有点中价值最大的是多少

solution：

每个格点可以贡献的范围显然是一个圆，令R = ceil(r)，将原网格扩充为N = n + 2*R，M = m + 2*R，那么，贡献的转移就像是二维空间上的卷积一样，将二维上的点和转移方式转换成一维，FFT解决

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#include<vector>#include<queue>#include<set>#include<map>#include<stack>#include<bitset>#include<ext/pb_ds/priority_queue.hpp>using namespace std;const int maxn = 3E6 + 20;typedef double DB;const DB eps = 1E-9;const DB PI = acos(-1);struct Virt{DB r,i; Virt(){}Virt(DB r,DB i): r(r),i(i){}Virt operator + (const Virt &b) {return Virt(r + b.r,i + b.i);}Virt operator - (const Virt &b) {return Virt(r - b.r,i - b.i);}Virt operator * (const Virt &b) {return Virt(r*b.r - i*b.i,r*b.i + i*b.r);}}A[maxn],B[maxn],C[maxn];int n,m,R,M,N;DB r;void Rader(Virt *F){int j = (N >> 1);for (int i = 1; i < N - 1; i++){if (i < j) swap(F[i],F[j]);int k = (N >> 1);while (j >= k) j -= k,k >>= 1;j += k;}}void FFT(Virt *F,int on){Rader(F); DB G = 2.00*PI*(DB)(on);for (int k = 2; k <= N; k <<= 1){Virt wn = Virt(cos(G / (DB)(k)),sin(G / (DB)(k)));for (int i = 0; i < N; i += k){Virt w = Virt(1,0);for (int j = i; j < i + (k >> 1); j++){Virt u = F[j],t = w * F[j + (k >> 1)];F[j] = u + t; F[j + (k >> 1)] = u - t;w = w * wn;}}}if (on == -1)for (int i = 0; i < N; i++) F[i].r /= (DB)(N);}int main(){while (scanf("%d%d%lf",&n,&m,&r) != EOF){R = ceil(r); M = m + 2*R; N = 1;while (N < (n + 2*R)*(m + 2*R)) N <<= 1;for (int i = 0; i < n; i++)for (int j = 0; j < m; j++){DB x; scanf("%lf",&x);A[i*M + j] = Virt(x,0);}for (int i = -R; i <= R; i++)for (int j = -R; j <= R; j++){DB dis = sqrt(i*i + j*j);if (r > dis)B[(i + R)*M + j + R] = Virt(1.00 / (dis + 1.00),0);}FFT(A,1); FFT(B,1);for (int i = 0; i < N; i++) C[i] = A[i] * B[i];FFT(C,-1); DB Ans = 0;for (int i = 0; i < n; i++)for (int j = 0; j < m; j++)Ans = max(Ans,C[(i + R)*M + j + R].r);printf("%.3f\n",Ans);for (int i = 0; i < N; i++) A[i] = B[i] = C[i] = Virt(0.00,0.00);}return 0;}

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