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From NOI2006, this is probably the first network stream in history.Remember how to build a maximum-weight closed sub-graph. Also remember that the network flow of the number of sides and points must not be easily determined!#include #include #include #include #include #define MAXV 55500#define Maxe 1000005#define INF 0X7FFFFFFFusing namespace Std;struct Edge{int v,f,nxt;}e[maxe];int nume=1,n,m,x,y,z;int G[MAXV],S=0,T,SUM=0,MAXCUT=0,DIS[MAXV];inline vo

" ...) ), then the problem becomes super piano ....1#include 2#include 3#include 4#include 5#include 6#include 7 #definePAU Putchar (")8 #defineENT Putchar (' \ n ')9 #defineCH for (int d=0;dTen #defineLson x->ch[0],l,m One #defineRson X->ch[1],m+1,r A using namespacestd; - Const intmaxn=500000+Ten, maxnode=3000000+Ten, maxm=1000000+Ten, maxt=maxn* -, inf=-1u>>1; - structted{intX,Y,W;TED*NXT;} adj[maxm],*fch[maxn],*ms=adj; the voidAddintXintYintW) { -

, D; Martix:: Martix () {memset(Mat,0,sizeof(MAT));} Martix Martix::operator* (ConstMartix b)Const{Martix ret; for(inti =0; I 2; ++i) { for(intj =0; J 2; ++J) { for(intK =0; K 2; ++K) {Ret.mat[i][j] + = ThisMAT[I][K] * B.mat[k][j]; RET.MAT[I][J]%= mod; } } }returnRET;} Martix Martix::operator+ (ConstMartix b)Const{Martix ret; for(inti =0; I 2; ++i) { for(intj =0; J 2; ++J) {Ret.mat[i][j] = ThisMAT[I][J] + b.mat[i][j]; RET.MAT[I][J]%= mod; } }returnRET;} martix Mart

How to use: sz (size) storage subtree size DP (deep) storage node depth fa (father) storage node Father HS (heavy son) storage node heavy son HF stores the top node of the current node's chain ID Storage node number nd (node) stores the current number Corresponding node Namespace Tree {#define MAXN 100001int SZ[MAXN], DP[MAXN], fa[maxn];int hs[maxn] = {0};void DFS1 (int x, int f, int dep) {D P[X] = DEP, fa[x] =

some 22 nonadjacent lattice, so that the weight of the largest ...Analysis:The most power value independent set problem, the grid diagram, so can black and white dyeing, s to Black point a capacity for the lattice weight of the edge, white point to t even a capacity for the lattice weight of the edge, the black point to the adjacent white point with a capacity of +∞ edge, and then the smallest cut is the minimum point weight coverage, The total weight minus the minimum cut is the answer ...Code

son to add that number, so we will be based on the depth of the odd couple, the time stamp into 2 arrays, the establishment of 2 segment tree/************************************************************************* > File name:cf225-e.cpp > Auth Or:alex > Mail: [email protected] > Created time:2015 April 21 Tuesday 21:11 04 seconds ******************************** ****************************************/#include #include #include #include #include #include #include #include #include #include

How far away?This problem test instructions is, given a tree, each side has a certain weight, q times asked, each time asked about the distance between two points. So you can use LCA to solve, first find the U, v two point of the LCA, and then calculate the distance value on it. Here the calculation method is, note the root node to any point of distance dis[], so that ans = dis[u] + dis[v]-2 * DIS[LCA (V, v)], this expression is relatively easy to understand://==================================

format:For each coach's name, output one line. If the name is correct and is first appeared, output "OK", if the name is wrong, output "wrong", if the name is correct but not first appear, output "REPEAT". (all without quotation marks)Ideas:It is said that the hash can be past, but I still obediently wrote the trie treeWe've built everyone's name into a trie tree.And then run a match.If there are no nodes on the trie and there are no matches, then the point is wrong.If the point to the name is

with three integersu,v andWwhere0≤w≤ Denoting an edge between u and v of Barricade cost w. Outputfor Each test cases, the output of the minimum wood cost. Sample Input14 2 4 1 3 4 Sample Output4Analysis: The shortest cut maximum flow can be obtained;Code:#include #include#include#include#include#include#include#includestring>#includeSet>#include#include#include#include#include#defineRep (I,m,n) for (i=m;i#defineRSP (It,s) for (Set#defineMoD 1000000007#defineINF 0x3f3f3f3f#defineVI vector

Hey hey, I wrote it before. I just suddenly saw it, so I posted it. KMP is used to search for the next [] array and the longest repeated string. Pku2406 # Include PKU 1961 # Include HDU 3746 cyclic nacklace View code # Include # Include Using Namespace STD; Const Int N = 100000 + 10 ; Char P [N]; Int NXT [N], M; Void Get_next () {NXT [ 0 ] =- 1 ; Int K =- 1 ; For ( Int I =

with revenue, each augmented path is actually the minimum traffic on the augmented roadso it's not the biggest payoffto run min cutans= Ownership Value and-min cutExplanation: Step 2: If you cut the same selection of this side, then the two students can only cut the edge of the text, otherwise it will be through the INF edge with the meeting point, does not meet the minimum cut#include #include#include#defineN 50101#defineM n*30#defineINF 2e9using namespacestd;intfront[n],to[m],

, where VI (1≤VI≤108) indicating the value of Pond I.Each of the last m lines contain-numbers A and B, which indicates that pond A and pond B is connected by a pipe.OutputFor each test case, output the sum of the value of all connected components consisting of odd number of ponds after removi Ng all the ponds connected and less than the pipes.Sample Input17 71 2 3 4 5 6 71 41 54 52 32 63 62 7Sample Output21HINTTest instructionsGive you a picture, and then ask for the number of points less than 2

is mx[], farthest right end position One intans=0; A intMain () { -scanf"%d%d",n,t); - intb; the inti,j; - for(i=1; i){ -scanf"%d%d",a,b); -mx[a]=Max (mx[a],b); + } - intlast=1, now=1; + intEd; A while(nowt) {//now is the length that is currently covered atEd=0; - for(i=last;i){ - if(mx[i]>ed) { -ed=mx[i];last=i; - } - } in if(ed+1==Now ) { -printf"-1\n"); to return 0; + } -Now=ed+1; ans++; the

sample ... TAT ... Because there are rings ...So we have to delete the point of the ring, and if the point on the ring is not selectable, then the point pointing at it cannot be selected ... What to do? Topological sort?But the topological sort can only judge if there is no ring, cannot judge whether the point is on the ring ... For example, the following picture ...But it's a coincidence that our chart is that all points pointing to the ring are not selectable ... So reverse all sides ... Topo

of investment 6, profit is 10, so the maximum gain of 4. "Scoring method" is not part of the subject, your program output only and our answer exactly the same to obtain full marks, otherwise do not score. Data size and convention 80%: N≤200,m≤1 000. 100% of the data: N≤5 000,m≤50 000,0≤ci≤100,0≤pi≤100.SOURCE Analysis:Bare Maximum right-closing sub-graph ...Each transit station as a negative power point, customer groups as a positive point, and then each customer to the transfer station to conne

mid ... TAT ...Code:1#include 2#include 3#include 4#include 5 //by Neighthorn6 #defineINF 0x3f3f3f3f7 #defineINF 0x3f3f3f3f3f3f3f3f8 using namespacestd;9 Ten Const intmaxn= $+5, maxm=100000+5; One A intn,m,s,t,cnt,sum,a[maxn],b[maxn],hd[maxn*2],fl[maxm],to[maxm],nxt[maxm],pos[maxn*2]; - Long LongDis[maxn][maxn],max; - theInlinevoidAddintSintXinty) { -fl[cnt]=s;to[cnt]=y;nxt[cnt]=hd[x];hd[x]=cnt++; -fl[cn

Topic Link: hdu_5963_ FriendTest instructionsChinese, not explainedExercisesTake a look at the sample, and you'll see if X is the node that wins, or whether the Benquan value associated with x is the same or odd.1#include 2 #defineF (I,A,B) for (int i=a;i3 using namespacestd;4typedefLong Longll;5 6 Const intn=80007;7 intt,n,m,g[n],nxt[n],v[n],w[n],ed;8 9 voidAdgintXintYintZ) {v[++ed]=y,w[ed]=z,nxt[ed]=g[x],

update the time to find all the direct son of U set to v OK, can be updated directly with F[V][CU], but if the nearest center of V is not CU, but is CV, then in the subtree of V to enumerate all CVs to find a minimum of F[V][CV], with f[v][cv]+k to update, Then why do you add K now? Since the CV is in the subtree of V, the center of U is not a CV but a CU, and it is a certainty that you have this CV built as a center, and you must add K.You can then use the Floyd to process the distance between

the list to store the position of all the puddings for each color, merge the linked list and color markers each time it is modified, record the link list length, and the list of violent modified lengths.However the writing hung up. After bowing to the puzzle, it was found that the original color of the pudding was not recorded, causing the stain to deviate.↓ Note the statement differences between lines 37 and 38. To determine whether to dye according to the primary color.1#include 2#include 3#i

() or listen (). This segment is discarded, but RST is not returned. The client tries to re-establish the connection request. Create a new socket: When a socket in the listening status receives this segment, a sub-socket will be created, and socket {}, tcpcb {} And pub {} will also be created. If an error occurs, the corresponding socket is removed and the memory is released through the flag. The TCP connection fails. If the cache queue is full, TCP considers an error to occur and all subseque