Hdu3715 2-sat + binary

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Determine the maximum number of conditions allowed by a 0/1 array (size: N. Conditions are given in arrays A, B, and C.

Tucao: Well, it's been a long time! The key is to abstract the question model (the above sentence ). Perform the second sat in the future: What are points and what are conditions !, Then pay attention to the details. This time, a small error occurred:

When there is no solution, I and I + 1 are in the same SCC, and I did not have + 2 each time! But ++! Silly... Success! I always thought that I had written an error in binary mode...

# Include <iostream> # include <cstdio> # include <stack> # include <vector> using namespace STD; const int INF = 0x3f3f3f3f; const int maxv = 402; // MaxE = 500000; int dfn [maxv]; int low [maxv]; int vis [maxv]; int SCC [maxv]; int INS [maxv]; stack <int> sta; int times = 0; int numb = 0; vector <int> E (maxv); void Tarjan (int u) {dfn [u] = low [u] = times ++; INS [u] = 1; Sta. push (U); For (INT I = 0; I <E [u]. size (); I ++) {int v = E [u] [I]; If (! Vis [v]) {vis [v] = 1; Tarjan (V); If (low [v] <low [u]) low [u] = low [v];} else if (INS [v] & dfn [v] <low [u]) low [u] = dfn [v];} If (low [u] = dfn [u]) {numb ++; int cur; do {cur = Sta. top (); Sta. pop (); INS [cur] = 0; SCC [cur] = numb;} while (cur! = U) ;}} int n, m; int A [10005], B [10005], C [10005]; void Init () {numb = times = 0; for (INT I = 0; I <maxv; I ++) {SCC [I] = INS [I] = dfn [I] = low [I] = vis [I] = 0; E [I]. clear () ;}} bool build_solve (int x) {Init (); For (INT I = 0; I <= x; I ++) {If (C [I] = 2) {e [2 * A [I] + 1]. push_back (2 * B [I]); e [2 * B [I] + 1]. push_back (2 * A [I]);} else if (C [I] = 1) {e [2 * A [I]. push_back (2 * B [I]); e [2 * B [I]. push_back (2 * A [I]); e [2 * A [I] + 1]. push_back (2 * B [I] + 1); E [2 * B [I] + 1]. push_back (2 * A [I] + 1);} else if (C [I] = 0) {e [2 * A [I]. push_back (2 * B [I] + 1); E [2 * B [I]. push_back (2 * A [I] + 1) ;}} for (INT I = 0; I <= 2 * n-1; I ++) {If (! Vis [I]) {vis [I] = 1; Tarjan (I) ;}} for (INT I = 0; I <= 2 * N-2; I + = 2) // start to write it as I ++ !!!! Wa to kneel! If (SCC [I] = SCC [I + 1]) return 0; return 1 ;}void readin () {for (INT I = 0; I <m; I ++) scanf ("% d", & A [I], & B [I], & C [I]);} int main () {int t; scanf ("% d", & T); While (t --) {scanf ("% d", & N, & M ); init (); readin (); int L = 0, r = m, mid; while (L + 1 <r) {mid = (L + r)/2; if (build_solve (MID) L = mid; else r = mid;} printf ("% d \ n", L + 1) ;}return 0 ;}