The largest clique high-connectivity component of ultraviolet A 11324 contraction point to construct a DAG

Question: The largest clique of the ultraviolet (a) 11324
Start at a certain point and find the maximum number of nodes that can be walked at a time

Analysis:
Tarjan calculates GCC, and then constructs a new graph. The new graph is a dag. For the Dag, use DP to find the longest path. The DP transition equation is DP [x] = size [x] + max (DP [y]). After the point is scaled down, the edge X to Y is displayed. You can simply perform a memory search, specific implementationCode

 

# Include <iostream> # include <cstring> # include <cstdio> # include <vector> using namespace STD; const int maxn = 1005; const int maxm = 50005; # define debug puts ("here"); int dfn [maxn], low [maxn], stack [maxn], Father [maxn], bcnt, top, depth; bool instack [maxn]; int po [maxn], Tol, n, m; int ID [maxn]; int DP [maxn]; int sum [maxn]; vector <int> VEC [maxn]; struct node {int y, next;} edge [maxm]; void add (int x, int y) {edge [++ tol]. y = y; edge [tol]. next = po [X]; po [x] = tol;} void DFS (int x) {// Recursive Implementation of Tarjan Algorithm Low [x] = dfn [x] = ++ depth; instack [x] = true; stack [++ top] = x; int y; for (INT I = po [X]; I; I = edge [I]. next) {Y = edge [I]. y; If (! Dfn [y]) {DFS (y); low [x] = min (low [X], low [y]);} else if (instack [y]) low [x] = min (low [X], dfn [y]);} If (low [x] = dfn [x]) {++ bcnt; do {Y = stack [top --]; instack [y] = false; father [y] = bcnt;} while (X! = Y) ;}} void Tarjan () {memset (low, 0, sizeof (low); memset (dfn, 0, sizeof (dfn )); top = bcnt = depth = 0; For (INT I = 1; I <= N; I ++) if (! Dfn [I]) DFS (I);} int F (INT X) {// evaluate the longest path of Dag if (DP [x]) return DP [X]; int ans = 0; For (INT I = 0; I <(INT) VEC [X]. size (); I ++) {// starting from all the edges of X, find the maximum path int y = VEC [x] [I]; ans = max (ANS, f (y); // transfer equation} DP [x] = ans + sum [X]; return DP [X];} void DAG () {memset (ID, 0, sizeof (ID); memset (sum, 0, sizeof (SUM); memset (DP, 0, sizeof (DP); For (INT I = 1; I <= N; I ++) VEC [I]. clear (); For (INT x = 1; x <= N; X ++) {// create a new graph for (Int J = po [X]; j = ed GE [J]. Next) {int y = edge [J]. Y; If (father [x]! = Father [y]) {VEC [Father [x]. push_back (father [y]); Id [Father [y] ++;} sum [Father [x] ++; // count the number of nodes of all source images contained by the node after each contraction} int ans = 0; For (INT I = 1; I <= bcnt; I ++) if (! Id [I]) ans = max (f (I), ANS); cout <ans <Endl;} int main () {freopen ("sum. in "," r ", stdin); int ncase; CIN> ncase; while (ncase --) {CIN> N> m; int X, Y; memset (PO, 0, sizeof (PO); Tol = 0; while (M --) {scanf ("% d", & X, & Y ); add (x, y);} Tarjan (); DAG ();} return 0 ;}