duckduckgo unblocked

Unblocked Project Continuation time limit:3000/1000ms (Java/other) Memory limit:32768/32768k (Java/other) total submission (s): Accepted s Ubmission (s): 37Problem description A province has finally built many roads since it has implemented many years of smooth engineering projects. But the road is not good, every time from one town to another town, there are many ways to choose, and some programmes are more than others to walk the distance is much sh

Unblocked Works continuedTime limit:3000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 36946 Accepted Submission (s): 13591Problem description a province since the implementation of many years of smooth engineering plans, finally built a lot of roads. But the road is not good, every time from one town to another town, there are many ways to choose, and some programmes are more than others to walk the distance is m

Unblocked Works continuedTime limit:3000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 53806 Accepted Submission (s): 20092Problem description a province since the implementation of many years of smooth engineering plans, finally built a lot of roads. But the road is not good, every time from one town to another town, there are many ways to choose, and some programmes are more than others to walk the distance is m

Smooth Project ContinuationTime limit:3000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 28356 Accepted Submission (s): 10275Problem DescriptionA province has built up a lot of roads since it implemented a project plan that has been unblocked for many years. Just a lot of road is not good, every time from a town to a town, there are many ways to choose the path, and some of the programmes are more than some of the

It took 1 days to write a script to detect whether an IP network is unblocked, but to detect whether the network is unblocked is easy to implement, but it is not good to add some restrictions. Copy Code code as follows: #!/bin/bash #################################################################################### #本脚本放于异机, the use of timed tasks to detect whether the network 192.168.10.231, i

said, BFS is a special dij. The core of DIJ is only three lines. The priority queue is then used. The following also describes how to use the pair object; AC code:/* /#include "algorithm" #include "iostream" #include "CString" #include "Cstdlib" #include "string" #include "Cstdio" # Include "vector" #include "cmath" #include "queue" using namespace std;typedef long long LL; #define MEMSET (x, y) memset (x, Y, sizeof (x)) #define MX 401/*/********************************************** #define ME

; the for(i=0; i) - for(j=0; j) Wu { - if(j==i) a[i][j]=0; About Elsea[i][j]=MAX; $ } - for(i=0; i) - { -Cin>>s>>e>>D; A if(D//There are many roads between two towns + { thea[s][e]=D; -a[e][s]=D; $ } the } thecin>>s>>e;//Start and End the //data read-in completed the Dijkstra (s); - if(Dis[e]==max) cout"-1"Endl; in ElsecoutEndl; the } the return 0

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1874Unblocked Works continuedTime limit:3000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 36359 Accepted Submission (s): 13355Problem description a province since the implementation of many years of smooth engineering plans, finally built a lot of roads. But the road is not good, every time from one town to another town, there are many ways to choose, and some programmes are more than others to walk the

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1232Analysis: The essence of the subject is a and check the set of problems, Ah, and check set is not very understanding, began to do not know what method to do. The problem of dynamic connectivity is also found in general use and solution. The key is to find out the number of isolated points, the number of roads to be built is the number of isolated points-1. There is the result of the output, you should pay attention to how to input.and 3 m

information, including start/end/Weight valuesintTol//number of edges, assigned 0 before adding edgevoidAddedge (intUintVintW) {edge[tol].u=u; EDGE[TOL].V=v; Edge[tol++].w=W;}//sort functions to sort edges from small to large by weightBOOLCMP (Edge A,edge b) {returna.wB.W;}intFindintx) { if(f[x]==-1)returnx; returnf[x]=find (F[x]);}//incoming number, returns the minimum spanning tree weight, if not connected return-1intKruskal (intN) {memset (F,-1,sizeof(F)); Sort (Edge,edge+tol,cmp); intCnt

http://acm.hdu.edu.cn/showproblem.php?pid=1875I just want to ask why it has been wrong, ask the great God of the way to teach!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!#include #include #include #include using namespace Std;int C,k,cou;Double shortest;struct p{int x;int y;}POINT[110];BOOL visit[110];struct graph{int la,lb;Double JL;}G[5050];Double Hpy (P a,p b) {return sqrt (POW (a.x-b.x,2) +pow (a.y-b.y,2));}int cmp (Graph a,graph b) {Return a.jl}void Kruskal () {Sort (g,g+k,cmp);int i;memset (visit

Title: Click to open link#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. Hdu OJ 1863 unblocked project

#include #include #include #include #include #include #include #include #include #include #include #include using namespace STD;structdata{intU,v;DoubleW;} e[ the];BOOLCMP (data A, data b) {returnA.WDoublex[ -+5],y[ -+5];intn,m,bin[ the];intFind (intx) {intS for(s=x;bin[s]>=0; s=bin[s]); while(s!=x) {intT=BIN[X]; Bin[x]=s; x=t; }returns;}voidUnion (intX1,intX2) {intF1=find (x1), F2=find (x2);intT=BIN[F1]+BIN[F2];if(Bin[f1]>bin[f2]) {bin[f1]=f2; bin[f2]=t; }Else{bin[f2]=f1;

Topic Links: http://acm.hdu.edu.cn/showproblem.php?pid=1874Thinking Analysis: The problem is given a graph, the starting point and the endpoint, the minimum distance from the starting point to the endpoint is required to find out;The shortest-circuit length from the starting point to all other points is calculated using the Dijkstra algorithm, if the shortest-circuit length is int_max, indicating that no path is connected from the starting point to the point;The code is as follows:#include #incl

Analysis: template problem, direct template can be.#include HDU ACM 1874 unblocked Project continued

Three levels of violence have been wrong for a long time ...It is a tragedy to think of the non-direction graph as a graph.#include #include #include #include #include #include #include #include #include #include #include #include #include using namespace STD;Const intn= $+5;Const intinf=100000;intG[n][n],n,d[n][n];voidFloyd () {intI,j,k; for(i=0; i for(j=0; jif(I==J) d[i][j]=0;Else if(g[i][j]==0) D[i][j]=inf;ElseD[I][J]=G[I][J];//printf ("%d%d:%d\n", I,j,d[i][j]);} for(k=0; k for(i=0; i for(j=0

Finding or thinking is the most important to understand the code of thought is easy to understand.#include nyoj608 unblocked project and check set

other n-1. - { - intbig=Long_max; - for(intj=1; j//The least right edge, and the corresponding point + { - if(!vis[j] low[j]big) + { Apos=J; atbig=Low[j]; - } - } -Ans+=big;//Long Statistical path -vis[pos]=1; - for(intj=1; j//update weights to each point in if(!vis[j]) low[j]=min (low[j],v[pos][j]); - } to returnans; + } - the * $ Panax Notoginseng intMain () - { theFreopen ("Input.txt","R

http://acm.hdu.edu.cn/showproblem.php?pid=1232and check Set#include hdu-1232 unblocked Project

= x, e[k].y = y, e[k].d =D; atk++; - } - - intGet_head (intx) - { - while(x! = fa[x]) x=Fa[x]; in returnx; - } to + BOOLUnion (intXinty) - { the intfa_x =get_head (x); * intFa_y =Get_head (y); $ if(Fa_x! =fa_y) {Panax NotoginsengFa[fa_x] =fa_y; - return true; the } + return false; A } the + intMain () - { $ //freopen ("a.in", "R", stdin); $ intN, M, a, B, D; - while(SCANF ("%d", N), N) - { theK =0; -m = N (n1) /2;Wuyi the for(int