Poj1904/zoj2470 King's Quest (Tarjan determine strongly connected components)

The time limit for this question in poj is 15000 Ms. Looking at all DT, it is at least scary ......

The question is, N boys and N girls tell you the numbers of girls that each boy prefers, and then give an initial match (this initial match is a complete match ), then find all possible complete matches and output them in ascending order. Of course, if it is violent (of course I have never tried it), 2000 boys and 2000 girls, up to 20 million targeted edges, will be quite disappointing ......

I read a report from zanniu and turned it into a strongly connected problem:

First, create a Graph Based on the given directed edge, and then add a reverse edge to the graph based on the final complete match (that is, link the girl to the boys' edge based on the complete match). In this graph, A strongly connected point must be a valid point. Sort them and output them.

I started to use the priority queue and thought it would be very fast. I ran 3800 + MS and changed it to sort, which is 3400 MS +. The following code is used:

# Include <cstdio> # include <cstring> # include <stack> # include <queue> # include <algorithm> using namespace STD; const int n = 4010; struct node {int val; bool operator <(const node & A) const {return. val <Val ;}}; struct edge {int S, E, next;} edge [300010]; int N, e_num, vis_num, CNT; int head [N], tim [N], low [N], instack [N], belong [N]; void addedge (int A, int B) {edge [e_num]. S = A; edge [e_num]. E = B; edge [e_num]. next = head [a]; head [] = E_num ++;} void getmap () {int I, j, M, X; e_num = 0; memset (Head,-1, sizeof (head )); for (I = 1; I <= N; I ++) {scanf ("% d", & M); For (j = 1; j <= m; j ++) {scanf ("% d", & X); addedge (I, n + x) ;}}for (I = 1; I <= N; I ++) {scanf ("% d", & X); addedge (n + X, I) ;}} stack <int> st; void Tarjan (INT X) {int I, j; Tim [x] = low [x] = ++ vis_num; ST. push (x); instack [x] = 1; for (I = head [X]; I! =-1; I = edge [I]. next) {int u = edge [I]. e; If (Tim [u] =-1) {Tarjan (U); If (low [u] <low [x]) low [x] = low [u];} else if (instack [u] & Tim [u] <low [x]) low [x] = Tim [u];} If (Tim [x] = low [x]) {CNT ++; do {J = ST. top (); ST. pop (); instack [J] = 0; belong [J] = CNT;} while (J! = X);}/* sort by sort, it will be faster void out (int A [], int Len) {sort (a + 1, A + 1 + Len ); printf ("% d", Len); For (Int J = 1; j <Len; ++ J) printf ("% d", a [J]); printf ("% d \ n", a [Len]);} */void solve () {int I, j, A [n]; CNT = vis_num = 0; memset (TIM,-1, sizeof (Tim); memset (low, 0, sizeof (low); memset (instack, 0, sizeof (instack )); for (I = 1; I <= 2 * n; I ++) {If (Tim [I] =-1) Tarjan (I) ;} node cur; priority_queue <node> q; for (I = 1; I <= N; I ++) {int num = 0; For (j = H EAD [I]; J! =-1; j = edge [J]. next) {int u = edge [J]. e; If (belong [u] = belong [I]) {cur. val = u-n; q. push (cur); num ++ ;}} printf ("% d", num); While (! Q. empty () {node v = Q. top (); q. pop (); printf ("% d", V. val);} puts ("");}/* use sort code for (I = 1; I <= N; I ++) {// traverse the edges in the graph, if the start point and end point are in the same connected component, the valid point is int num = 0; For (j = head [I]; J! =-1; j = edge [J]. next) {int u = edge [J]. e; If (belong [I] = belong [u]) {A [++ num] = u-n ;}} out (a, num );} * // puts (""); note that the zoj output requires an empty line more than the poj output.} int main () {While (~ Scanf ("% d", & N) {getmap (); solve ();} return 0 ;}