forest engraving

We will give you a detailed analysis of photoshop software users and share some tutorials on shooting and post-processing in the dark forest.Tutorial sharing:First look at the effect:Exercise material:We shot it in a small and small forest at the corner of a road. It rained at around five o'clock P.M. that morning, and the light was particularly dark, and many ants.Equipment: Canon 5D3 + 50 1.4Post-tool: PS + LRMakeup is a model's own speciali

The principle of isolation Forest algorithm introduced in this article is described in my blog: Isolation Forest anomaly detection algorithm principle, we only introduce the detailed code implementation process in this article.1, the design and implementation of ItreeFirst, we refer to the construction pseudocode of Itree in the original paper:Write a picture description here1.1 Designing the data structure

Python3 learn the API of using random forest classifier gradient to promote decision tree classification and compare them with the single decision tree prediction resultsAttached to my git, please refer to my other classifier code: https://github.com/linyi0604/MachineLearning1 ImportPandas as PD2 fromSklearn.cross_validationImportTrain_test_split3 fromSklearn.feature_extractionImportDictvectorizer4 fromSklearn.treeImportDecisiontreeclassifier5 fro

A forest logo implemented with CSS and HTML The HTML code is as follows: The CSS code is as follows: * {margin:0; padding:0;}. Logo {height:300px; width:260px; margin:150px Auto; Position:relative;} #tree {border-bottom:300px solid #063; border-left:130px solid Transparent; border-right:130px solid Transparent; Position:absolute; left:0; top:0; height:0; width:0;} #trunk {height:180px; width:32px; Background:

First for the query operation is the bare cot qaqDFS builds the Chairman tree on the tree.For the connection operation, we found that we did not deleteSo we can do heuristic merging, and each time a small tree is flattened into a large tree and refactored.After writing, the first and second points of the puzzle のreAnd then re-wrote it again for a (don't know why, did you write down the first time?)#include 　　Bzoj 3123 SDOI2013 Forest

. Although today did not meet, want that temple cloud lazy sweep, eyebrow dai micro cu, tired new song old harp, tired of black and white keys. Do not want to contest, like spring to Xie Yellowstone plum, thin loss of waist, such as after autumn after the mountain Longan, but in today's mirror, feeling old sorrow, loss of taste. This Lin two still in sentimental, suddenly see own door tight, although is star Speed rescue, also difficult to solve near fire. Unfortunately, the city broke, but was

To find the shortest path to other points, the conditions of the topic become u->v not backtrack when and only when D[u]>d[v]. The DAG graph is then built according to this condition, running DP statistics scheme number, dp[u] = SUM (Dp[v]).#include using namespacestd;Const intMAXN =1001, MAXM =2002;structedge{intv,w,nxt;};#definePB push_backVectorEdges;vectorint>G[MAXN];intHEAD[MAXN];intD[MAXN];voidAddedge (intUintVintW) {edges. PB ({V,w,head[u]}); Head[u]= Edges.size ()-1;}voidinit () {mems

Title Link: Http://codeforces.com/problemset/problem/501/CTest instructionsIdeas:Code:#include #include#include#include#includeSet>#include#include#include#include#defineLson (i#defineRson (lson|1)using namespaceStd;typedefLong LongLL;Const intN =1000007;Const intINF =0x7fffffff;structnode{intidx; intde; intVal;} Dt[n];structans{intA, B;} Ans[n];intN, Vis[n], Cnt_ans;voidBFs () {Queueint>que; for(inti =0; I ) { if(Dt[i].de = =1) {Vis[i]=1; Que.push (i); } } while(!Que.em

] + dis[t][v[t][i]) {D[v[t][i]] = D[t] + dis[t][v[t][i]];if(!vis[v[t][i]]) {Vis[v[t][i]] =1; Q.push (V[t][i]); } } } }}intDfsintcur) {if(cur = =2)return 1;if(Dp[cur])returnDp[cur]; for(inti =0; I if(D[cur] > D[v[cur][i]) {Dp[cur] + = DFS (v[cur][i]); } }returnDp[cur];}intMain () { while(scanf("%d", n)! = EOF N) {scanf("%d", m); Init (); SPFA ();printf("%d\n", DFS (1)); }return 0;} Copyright NOTICE: This article for Bo Master original article,

, because it is the encapsulation system, to ensure the fewest hardware devices.650) this.width=650; "title=" 15.png "src=" http://s3.51cto.com/wyfs02/M01/56/BF/wKiom1SL-R7hYTcbAAE8qlBnK2w716.jpg "alt=" Wkiom1sl-r7hytcbaae8qlbnk2w716.jpg "/>16. In the hardware list, remove "floppy disk drive", "USB controller", "sound card", "Virtual printer", then click "OK" and go back to "15." Click "Finish".650) this.width=650; "title=" 16.png "src=" http://s3.51cto.com/wyfs02/M00/56/BC/wKioL1SL-cfQe2I3AAG_p

(edge+1, edge+1+m, CMP);inttot=0; for(intI=1; im; i++) {if(edge[i].tag==1)Continue;int x=edge[i].x,y=edge[i].y;if(Find_root (x)!=find_root (y)) {val[i+n]=edge[i].b;Link(x, I+n),Link(I+n,y); }Else{intT=query (x,y);if(Edge[i].bx, T), Cut (T,edge[t-n].y);Link(x, I+n),Link(I+n,y); } }intTmp=query (1, n);if(tmp!=-1) {ans=min (ans,edge[i].a+val[tmp]); } }if(Ans==inf)printf(" -1\n");Else printf("%d\ n", ans);} Copyright NOTICE: This article for Bo Master original article, without Bo Master pe

idea: Similar to a topological sorting problem, according to the degree of 1 points, it is connected to the number of edges is different or and this rule, directly to the topological sort. Each time the resulting node is reduced by one, and xor the node he or she is connected to.Code:#include Codeforces Round #285 (Div. 2) C-misha and Forest

Title Link: Http://codeforces.com/contest/501/problem/CTest instructions: Gives the sum of the degrees of each vertex of the graph and the XOR values of the adjacent vertices, and the number of edges and edges of the graph:Analysis: Find the point of a degree, the corresponding XOR value is the adjacent vertex (Code:#include Codeforces #285 Div.2 C. Misha and Forest

digital wave quickly brought about the address depletion crisis. The main purpose of IPv6 is to increase IPv4 address capacity, but it also makes improvements based on IPv4 experience and technological advances in the new era, such as avoiding fragmentation, for example, to cancel checksum (due to the widespread use of high-level TCP protocol ). Network protocols are not technically complex. More considerations are:Policy. The IP protocol is "Best Effort", and IP transmission is unreliable. H

achieve the smooth implementation of the relay, we further go deepARPAndRip/BGP. All three Protocols assist in IP transmission. ARP allows each computer and router to know the ing between the IP address and the MAC address in their LAN, so as to smoothly encapsulate the IP packet to the frame. The RIP Protocol can generate a reasonable routing table in the autonomous system. The BGP protocol can generate a routing table outside the autonomous system. Throughout the entire process, we focus on

TCP protocol specifies that when the receiver receives unordered fragmentsSend ACK repeatedly. For example, when receiving the unordered Part 9, the receiver needs to reply to ack. The reply number is 8 (7 + 1 ). After that, if the recipient continues to receive unordered segments (segments whose serial number is not 8), ACK = 8 will be sent again. When the sender receives 3 ACK = 8 replies, the sender infers that segment 8 is lost. Even if the timer of Segment 8 has not timed out, the sender w

NOJ1119 Xianlin dingshan amusement park, noj1119 forest dingshanQuestion If a directed and untitled graph is given, check whether a ring exists in the graph.Ideas Perform preprocessing in the form of Floyd. Eg [I] [j] not only indicates that an edge is connected to I and j, but also indicates that there is an I-to-j Path.In this way, we traverse all the situations. If we find that both the positive and the opposite are acceptable, there will be loops.

HDU 4941 Magical Forest [discretization] [map], hdu4941 Question link: http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 4941 Here are 10 ^ 5 vertices. Each vertex has a numerical value. The xy coordinate of the vertex is 0 ~ 10 ^ 9, points exist in the matrix. Then, a 10 ^ 5 operation is given. 1 indicates the exchange row, 2 indicates the exchange column, and 3 indicates the value where the query coordinate is the xy point. A large amount of data... A

issuedMessage complaint: record id, title, content, release time, submission of household, reply content, reply time, reply personAdmin: username, Login password, admin category, name, gender, Contact PhonePhoto Name: 1 System LoginPhoto name: 2 Super Admin Add other adminPhoto Name: 3 Other administrator information managementPhoto name: 4 Information Administrator login Management BuildingPhoto Name: 5 Add employee informationPhoto name: 6 Employee information ManagementPhoto Name: 7 Add busi

Label: style blog HTTP Io color OS AR for SP Question: There are N * m grids (n, m Solution: the grid is too large, but K is relatively small. So consider discretization and separate the rows. A maximum of 10 ^ 5 rows can be saved in map. When nowr [I] = J, the current row I is the original row J, and nowc indicates the column R [] indicates whether the row has a value, and CNTC [] indicates a column. Then, you can switch the above items directly during the exchange. When querying,