HDU 4612 Warm up (double connected component indent + tree diameter)

Idea: Strong connected component contraction point, build a new tree, and then find the longest diameter of the tree, and then add a side can remove the number of bridges, is the length of the diameter.

Tree diameter length of the method: two times BFS can be begged, for the first time to find a point u, and then the BFS search to the last point V, must be a diameter of the end (proof withheld), the second point V for the beginning of the BFS, the last point to find out, is the diameter of another endpoint, Record depth is the length of our request. I am here to use the Bfs+dfs, is the same, less open a deep array, save space bar ...

In fact, I do not want to start from the beginning, I thought that a leaf node can do the end point, hand it to WA, was also curious to feel right, the result of a random painting, it is wrong ...

Note: There is a heavy edge in this problem, we can go back to the Father node by the heavy side, but not through the father side back to the Father node, so the original way to mark the father is not possible, here to mark the edge, in the edge structure to add a vis variable, to mark whether this side has been visited, So the heavy side will be able to access the normal.

The problem of data volume is stronger, easy to explode the stack, the usefulness of STL is not strong, so try to use the array simulation, the array can open to a larger space.

Feelings: Feel this piece of the topic really love error, this problem is WA a good many times before, feel OJ don't love me ...

#include <iostream>#include<cstdio>#include<cstring>#include<vector>using namespacestd;#defineN 200020#defineM 2000020structedge{intTo,nxt,vis;} EDGE[M];intHead[n],dfn[n],low[n],vis[n],sta[n],id[n];intTot,all,top,scc,bridge;voidAddedge (intAintBintflag) {edge[tot].to=b; EDGE[TOT].NXT=Head[a]; Edge[tot].vis=Flag; Head[a]= tot++;}voidTarjan (intu) {Vis[u]=1; Dfn[u]= Low[u] = + +All ; Sta[top++] =u;  for(inti = Head[u]; I! =-1; i =edge[i].nxt) {        if(Edge[i].vis)Continue; intv =edge[i].to; Edge[i].vis= edge[i^1].vis =1; if(!Dfn[v])            {Tarjan (v); Low[u]=min (low[u],low[v]); if(Low[v] > Dfn[u]) bridge++; }        Else if(Vis[v]) low[u] =min (low[u],dfn[v]); }    if(Low[u] = =Dfn[u]) {SCC++; intnum;  Do{num= sta[--top]; Id[num]=SCC; Vis[num]=0; }         while(U! =num); }}vector<int>Tree[n];intTreevis[n],maxdeep;voidDfsintUintDeep ) {Treevis[u]=1; Maxdeep=Max (deep,maxdeep); intLen =tree[u].size ();  for(inti =0; i < Len; i++)    {        intv =Tree[u][i]; if(!Treevis[v]) {DFS (V,deep+1); }    }}voidinit () {memset (Vis,0,sizeof(VIS)); memset (DFN,0,sizeof(VIS)); memset (ID,0,sizeof(ID)); memset (Low,0,sizeof(low)); All=0; Top=0; SCC=0; Bridge=0;}intQue[n];intGetstart () {intfront,rear; Front= Rear =0; Que[rear++] =1; treevis[1] =1;  while(Front! =rear) {        intnow = que[front++]; intLen =tree[now].size ();  for(inti =0; i < Len; i++)        {            intNXT =Tree[now][i]; if(TREEVIS[NXT])Continue; TREEVIS[NXT]=1; Que[rear++] =NXT; }    }    returnque[--rear];}intMain () {intn,m,a,b;  while(~SCANF ("%d%d",&n,&m)) {if(!n &&!m) Break; memset (Head,-1,sizeof(head)); Tot=0;  for(inti =0; I < m; i++) {scanf ("%d%d",&a,&b); Addedge (A, B,0); Addedge (B,a,0);        } init (); Tarjan (1);  for(inti =1; I <= SCC; i++) {tree[i].clear (); }         for(intU =1; U <= N; u++)        {             for(inti = Head[u]; I! =-1; i =edge[i].nxt) {                intv =edge[i].to; if(Id[u]! =Id[v]) {                    intUU =Id[u]; intVV =Id[v];                    Tree[uu].push_back (VV);                Tree[vv].push_back (UU); } }} maxdeep=0; memset (Treevis,0,sizeof(Treevis)); intStart =Getstart (); memset (Treevis,0,sizeof(Treevis)); DFS (Start,1); printf ("%d\n", scc-maxdeep); }    return 0;}

HDU 4612 Warm up (double connected component indent + tree diameter)