clique studios

{ $ returnX y; - } - intnodecmp (node x, node Y) - { A returnx.x y.x; + } themapint,int>MP; - intCH[MAXN]; $ intMain () { them =0 ; thescanf"%d",n); the intTM; the for(inti =1; I ) - { inscanf"%d%d",qu[i].x,TM); theQU[I].L = qu[i].x-TM; theQU[I].R = qu[i].x +TM; Aboutch[m+1] =QU[I].L; thech[m+2] =QU[I].R; theM + =2; the } +Sort (qu+1, qu+1+n,nodecmp); -Sort (ch+1, ch+1+m,cmp); thech[0] = -1e9- +;Bayitt =0 ; the for(inti =1; I ) the { - if(Ch[i]!

Title Link: cf-r296-d2-dMain topicA special graph, some points on the axis, each point has a coordinate X, there is a weight of W, two points (I, J) have an edge between and only if | xi-xj| >= Wi + Wj.Ask for the largest regiment of this figure.The number of points in the graph N Problem analysisTwo points right between meet Xj-xi >= Wi + Wj (xi Sort by coordinates x from small to general points. Use f[i] to indicate the maximum group size of the first I point.Then f[i] = max (f[k]) + 1 (k This

(inti =1; I 0; for(intU =0; U ){ intV0 = Sccno[u]; wei[v0]++; for(inti = Head[u]; ~i; i =Nxt[i]) { intV1 =Sccno[to[i]]; if(V0! = v1) G[v0].push_back (v1), deg[v1]++; } }}intDP[MAXN];inttopo () {memset (DP,-1,sizeof(DP)); Queueint>Q; for(inti =1; I ){ if(!deg[i]) Q.push (i), dp[i] =Wei[i]; } while(Q.size ()) {intU =Q.front (); Q.pop (); for(inti =0; I int) G[u].size (); i++){ intv =G[u][i]; DP[V]= Max (dp[v],dp[u]+Wei[v]); if(--deg[v] = =0) Q.push (v

topological orderBayi intans=0; the while(!que.empty ()) the { - intsiz=que.size (); - for(intI=0; i//all nodes with a 0 entry level the { the intt=Que.front (); Que.pop_front (); theAns=max (ans,dp[t]+num[t]); the for(intj=0; J//each side that starts with T - { the intD=G[t][j]; ther[d]--; the if(!R[d]) que.push_back (d);94Dp[d]=max (dp[d],dp[t]+num[t]); the } the } th

]; Stackint>SintLinklow[n], Pre[n], head[n], sccno[n], num[n], dp[n];intTot, Dfs_clock, scc_cnt, N, M;BOOLLink[n][n];voidDfsintu) {linklow[u] = pre[u] = ++dfs_clock; S.push (U); for(inti = Head[u]; I! =-1; i = e[i].next) {intv = e[i].to;if(!pre[v]) {DFS (v); Linklow[u] = min (Linklow[u], linklow[v]); }Else if(!sccno[v]) {Linklow[u] = min (Linklow[u], pre[v]); } }if(Pre[u] = = Linklow[u]) {scc_cnt++; NUM[SCC_CNT] =0; while(1) {intx = S.top (); S.pop (

1. Title Description: Click to open the link2. Problem-Solving ideas: The game feels this problem should be more difficult, who knows to read someone else's code after found that they really are this paper tiger frightened = =. Assuming the point xi>xj, then the absolute value symbol can be removed, i.e. XI-XJ≥WI+WJ. XI-WI≥XJ+WJ can be obtained by moving items. In this case, in fact, we have determined a graph of the relationship, the topic into the number of nodes to find the most of the graph.

Topic Link: Click to open the linkTest instructionsN points on a given axis.The following n lines are two numbers per line XI, WI represents point and point weights.For any two points u, vIf Dis (u,v) >= u_w+v_w, one side can be built between the two points. (In the other words if the distance between two points is greater than two points of weight and can be built side)Find a maximum regiment and output the maximum number of points.Actually, for a weighted point, we can think of it as an interv

Question: given an adjacent matrix, find the largest group. Solution: direct AC. The Code is as follows: #include #include #include #include #include using namespace std;int N, mp[55][55];int ret, st[55], cnt[55];void dfs(int x, int num) { int

There is a direction graph G to find the node set with the largest number of nodes where any 2 points u and v either u can be to V or V can be to u or to each other to reach First, each of the strongly connected components is accessible, and the

Great group. That is, to find a maximum point set, so that any two vertices in the vertex set U, V exists at least U-> V, or V-> U path. In this way, find all the connected components, and then the entire graph becomes a non-cycle graph, reducing

For a directed graph G, finding a subgraph requires that any two of them have at least one side to be reachable, The maximum number of vertices contained in this subgraph. First, find the SCC contraction point to create a graph. The weight of each

factor chart. Psi represents probability or conditional probability. Factor graphs and PSI function notation are commonly used in Machine Learning paper. Potential Function explanation: Note: At the request of some friends, let's talk about the potential function. In this case, we can regard the potential function as the normalized decomposition factor of the joint probability density. The scope of the potential function is maxima clique ). Every

generally called energyfunction, defined as the sum of all clique-potential on the MRF. T is called temperature, generally take 1. What is Click-potential? That is, in the corresponding graph of MRF, each clique corresponds to a function called clique-potential. This joint probability form is also called gibbsdistribution. The Hammersley and Clifford theorem exp

, that is, all y1 of the molecule,.. Yn sum to get . U (Y1,.. Yn) is generally referred to as the energy function, defined as the sum of all clique-potential on the MRF. T is called temperature, generally takes 1. What is Click-potential? In the graph of MRF, each clique corresponds to a function called clique-potential. This joint probability form is also calle

. Clustering can be determined by looking for high-density areas in the new space. Wavelet transforms have the following advantages for clustering:Provides no guidance for clustering. It uses cap-shaped filtering, emphasizing point-dense areas, while ignoring weaker information outside the dense zone. Thus, the dense area in the original feature space becomes the attraction point of the nearby point, and the farther point becomes the restraining point. This means that the clustering of the data

Graph can be regarded as the association probability distribution formula of all elements in Y.Into multiple potential functions (potentialFunction), where each potential function acts on the adjacent vertex pair Yi and Y (I + 1 ). The purpose of normalization is to ensure that the product of the potential function is random variables in G.Effective probability distribution on the vertex. Note: At the request of some friends, let's talk about the potential function. In this case, we can regar

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