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Question: Given n points, m points have directed edges. Divide points into several sets so that any 2 points in each set are not reachable (a set can only store one point) The question must be divided into several sets at least. If no ring exists, this question is to find the longest chain of the directed graph, and run bfs In the topological order. However, when a ring exists, the point of point x is reduced to a new point x, and the point of point x is the number of vertices corresponding to t

Question link: Click the open link Question: Given n vertices, m records have no direction edge and Edge Weight (no duplicate edge) Find two shortest paths that do not overlap from 1-N points (the two paths must be the shortest path) First, we ran with the cost. Then, the shortest path is optimized, and, and all edges in the shortest path are deleted. Then you can actually run the network stream, and, at the beginning, it is still mle. Later, we removed the from the adjacent table ,, # Include

test case, print one line saying "to get from XX to YY takes N Knight moves."E2 e4//Start pointA1 B2B2 C3A1 H8A1 H7H8 A1B1 C3F6 f6Sample OutputTo get from E2 to E4 takes 2 Knight moves.To get from A1 to B2 takes 4 knight moves.To get from B2 to C3 takes 2 knight moves.To get from A1 to H8 takes 6 knight moves.To get from A1 to H7 takes 5 knight moves.To get from H8 to A1 takes 6 knight moves.To get from B1 to C3 takes 1 Knight moves.To get from F6 to F6 takes 0 knight moves.1#include 2#include

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[u]= tot++;}intMaxx, POS;voidBFsintP//wide search to find the farthest point from P{Maxx=0; memset (Vis,false,sizeof(VIS)); QueueQ; PII cur, NEX; Cur.first= P; Cur.second =0; VIS[P]=true; Q.push (cur); while(!Q.empty ()) {cur=Q.front (); Q.pop (); for(inti = Head[cur.first]; I! =-1; i =Edge[i].next) { intv =edge[i].to; if(Vis[v])Continue; VIS[V]=true; Nex.first= V; Nex.second = Cur.second +1; if(Maxx Nex.second) {Maxx= Nex.se

;typedef pairint>Node;#defineFi first#defineSe Secondll D[MAXN];voidDijkstraintS,ll P) {memset (d,0x7f,sizeof(d)); Priority_queueQ; Q.push (Node (d[s)=p,s)); while(Q.size ()) {Node x=q.top (); Q.pop (); intU =x.se; if(D[u]! = x.fi)Continue; ll T= (Tp[u]? (1+d[u]):((d[u]+ -)/ ++D[u])); for(inti = Head[u]; ~i; i =Nxt[i]) { intv =To[i]; if(D[v] >t) {Q.push (Node (d[v)=t,v)); }}}}ll Cost (intUintv) { returnTP[V]?1:((d[u]+ +)/ -);}voidFindpath (intSinte) { intU =s; while(U! =e) {p

allocated from the large pool. However, in many business systemsIf this parameter is not set, the parallel operation still uses the memory of the shared pool. We will discuss this later. Next, we can see that the PMON processes cannot obtain latch, and the system will no longer work at this time. Obviously, the root cause of this problem is the previous one.ORA-04031 error, here we need to analyze the trace to confirm why the process reports an error. The content of this trace file (xx1_p485_28

Descriptionjasio is a three-year-old boy, he likes to play toys, he has n different toys, they are placed on a very high shelf so jasio can't get them. In order to have enough space in his room, there will be no more than K toys on the floor at any moment. Jasio playing with toys on the floor. Jasio's mother stayed with his son in the room. When Jasio wants to play with the other toys on the floor, he will take it himself, if he wants to play the toy on the shelf, his mother will help him to tak

- for(i=0; i) to { + if(S1 (1i)) - Continue; the intNEX=S1 | (1i); *Dp[s1][s2]=min (Dp[s1][s2],max (Dfs (NEX,S2), DFS (nex,s2^ (11); $ }Panax Notoginseng returnDP[S1][S2]; - } the + intMain () A { the inti,j,k; + while(SCANF ("%d%d", m,n)! =EOF) - { $ if(n==0 m==0) $ Break; -Memset (P,0,sizeof(P)); -memset (Dp,inf,sizeof(DP))

", "w", stdout); * $ using namespacestd;Panax Notoginseng #defineN 10100 - //n is the maximum number of points the #defineM 50100 + //m is the maximum number of sides A intN, M;//N M is the number of points and sides the + structedge{ - int from, to, NEX; $ BOOLSign//whether it is a bridge $}edge[m1]; - intHead[n], edgenum; - voidAddintUintV) {//the beginning and end of the edge theEdge E={u, V, Head[u],false}; -Edge[edgenum] =E;WuyiHead[u]

(x) (x) (-X) - #defineRead () freopen ("A.txt", "R", stdin) the #defineWrite () freopen ("Dout.txt", "w", stdout); * $ using namespacestd;Panax Notoginseng #defineN 20100 - //n is the maximum number of points the #defineM 50100 + //m is the maximum number of sides A intN, M;//N M is the number of points and sides the + structedge{ - int from, to, NEX; $ BOOLSign//whether it is a bridge $}edge[m1]; - intHead[n], edgenum; - voidAddintUintV)

points and sides the + structedge{ - int from, to, NEX; $ BOOLSign//whether it is a bridge $}edge[m1]; - intHead[n], edgenum; - voidAddintUintV) {//the beginning and end of the edge theEdge E={u, V, Head[u],false}; -Edge[edgenum] =E;WuyiHead[u] = edgenum++; the } - Wu intDfn[n], Low[n], Stack[n], top, time;//Low[u] is the dfn[v of the point set {U-point and subtree in the root of the U-point (all reverse arcs) that can point to (the neares

One)+Ten]; - intP[MAXN]; the Charstr[ -]; + intm, N; A the intDFS (intS1,intS2) { + if(DP[S1][S2]! = INF)returnDP[S1][S2]; - intCNT =0; $ for(intI=1; ii) { $ if((P[i] S1) = = s2) cnt++; - } - if(CNT 1) { the returnDP[S1][S2] =0; - }Wuyi for(intI=0; ii) { the if(S1 (1Continue; - intNEX = S1 | (1i); WuDP[S1][S2] = min (Dp[s1][s2], Max (DFS (NEX, S2), DFS (

the tag array, because it is the minimum energy consumption, so the state of the mark is the current position and the energy consumed at this location, so you can use the tag array to store energy, if the current position state consumes the total energy is less than the location Update tag Array, code example #include#include #include #include #include #define MAXH 1000+100#define INF 10000using namespace Std;typedef struct NODE{int x,y,step;friend bool Operator {Return n2.step}}node;int dir[8]