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import numpy as np | |
import os | |
import cv2 as cv | |
from glob import glob | |
from scipy.spatial.transform import Rotation as Rot | |
from shutil import copytree | |
scans = ['106'] | |
# scans = [] |
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import pyrender | |
import os | |
from pyrender import PerspectiveCamera,\ | |
DirectionalLight, SpotLight, PointLight,\ | |
MetallicRoughnessMaterial,\ | |
Primitive, Mesh, Node, Scene,\ | |
Viewer, OffscreenRenderer, RenderFlags | |
from pyrender.constants import RenderFlags | |
import trimesh |
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class SineLayer(nn.Module): | |
# See paper sec. 3.2, final paragraph, and supplement Sec. 1.5 for discussion of omega_0. | |
# If is_first=True, os a frequency factor which simply multiplies the activations before the | |
# nonlinearity. Different simega_0 signals may require different omega_0 in the first layer - this is a | |
# hyperparameter. | |
# If is_first=False, then the weights will be divided by omega_0 so as to keep the magnitude of | |
# activations constant, but boost gradients to the weight matrix (see supplement Sec. 1.5) |
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from matplotlib import pyplot as plt | |
import numpy as np | |
n_samples = 128 | |
# pdf of logistic distribution | |
def pdf(x, std): | |
return 0.25 / std / np.cosh(0.5 * x / std)**2 |
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import numpy as np | |
import cv2 as cv | |
import os | |
from glob import glob | |
from scipy.io import loadmat | |
import trimesh | |
def clean_points_by_mask(points, scan): | |
cameras = np.load('/home/aska/Projects/NeuS/public_data/dtu_scan{}/cameras_sphere.npz'.format(scan)) | |
mask_lis = sorted(glob('/home/aska/Projects/NeuS/public_data/dtu_scan{}/mask/*.png'.format(scan))) |
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Size--> 5 | |
0 1 2 3 4 | |
5 6 7 8 9 | |
10 11 12 13 14 | |
15 16 17 18 19 | |
20 21 22 23 24 | |
X--> 1 | |
0 X 2 3 4 | |
5 6 7 8 9 | |
10 11 12 13 14 |
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#include <cstdio> | |
#include <cstring> | |
#include <algorithm> | |
#include <queue> | |
using namespace std; | |
queue <int> que; | |
double p[26],g[50][50],f[1 << 9][50]; | |
int i,j,k,l,m,n,L,t,cl,top,st,T,len; | |
int nex[50][26],fail[50],vis[50]; | |
int done[50],sing[50]; |
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#include <cstdio> | |
#include <cstring> | |
#include <algorithm> | |
#include <vector> | |
#include <iostream> | |
#include <map> | |
#define intl long long | |
#define maxn 300100 | |
using namespace std; |
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2014-2-26 | |
今天的考试没上一百分 | |
T1:考场上想到正解 并打出 但是一开始只有20分:首先我程序的常数太大 一个点要跑2秒 再次题目的数据范围明明是50W而不是题目所说的10W。。。不过总而言之还是自己的算法不够优秀,虽然主席树和树状数组的复杂度都是logn的,但是主席树的常数还是太大了(没有仔细分析题目性质的后果,本来可以用树状数组为什么用主席树?) | |
于是:“在时间充裕的情况下,能不能想出更优秀的算法?” | |
T2:这道题以前做过类似的题目(BZOJ 1003),打完了以后觉得没有什么萎点,于是去想T1,后来只对了一个点,下午又调了好久才知道错误的原因:对set的熟练程度不够。 | |
其实我知道set不允许重复元素的存在,但这个理解一直停留在表面:以为set<int> 里的a,b只要a != b就不是重复元素,后面才知道如果写了cmp函数的话。。set判重就靠cmp了。。。也就是说如果cmp(a,b)和cmp(b,a)都是false的话 就被set当成重复元素 然后就只会保留一个。。。于是嗯。。。就Wa了9个点。把set的cmp加上第二关键字就过去了。 | |
幸亏有这次错误,要不然我都不知道以后什么时候会在某个大考试犯它。。。 | |
T3:这一题没做,都去做前两题去了,其实这道题可以玩出很多分,提交答案题嘛。。 | |
不过如果有界面的话这道题会非常好做(就是一个推箱子游戏) | |
向GY学习如何打界面! |
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#include <cstdio> | |
#include <cstring> | |
#include <algorithm> | |
#define N 2000 | |
#define maxn 1000100 | |
#define intl long long | |
using namespace std; | |
int i,j,k,l,m,n,T,v[maxn]; | |
intl Mo,fac[maxn + 4010],ni[maxn + 4010],g[4010],f[4010],s[2010],js,ans; |
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