HDU 3715 Go deeper two points + 2-sat

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Topic:

http://acm.hdu.edu.cn/showproblem.php?pid=3715

Test instructions

Given a recursive pseudo-code, ask to perform the deepest number of recursion for this pseudo-code

Ideas:

Two-minute enumeration of the answer with 2-SAT to determine whether it is feasible. The concrete map is as follows: if c[i] = = 0, then a[i] or b[i], if c[i] = = 1, then (A[i] and b[i]) or (~a[i] and ~b[i]), if c[i] = = 2, then not (A[i] and b[i]), and then strongly connected Deflation point determine if I and ~i are within the same ring

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <cmath>using namespace STD;Const intN =410;Const DoubleEPS =1e-8;structedge{intTo, next;} g[n*n*2];intCNT, Head[n], cnt1, head1[n];intDfn[n], Low[n], scc[n], St[n], top, num, idx;intA[n*n], B[n*n], c[n*n];BOOLVis[n];intN, M;voidAdd_edge (intVintu) {g[cnt].to = u, G[cnt].next = Head[v], head[v] = cnt++;}voidInit () {memset(Head,-1,sizeofHead);memset(DFN,-1,sizeofDFN);memset(Vis,0,sizeofVIS); top = num = IDX = CNT =0;}voidTarjan (intV) {Dfn[v] = low[v] = ++idx; VIS[V] =true, st[top++] = v;intU for(inti = head[v]; I! =-1; i = g[i].next) {u = g[i].to;if(Dfn[u] = =-1) {Tarjan (U);        Low[v] = min (Low[v], low[u]); }Else if(Vis[u]) low[v] = min (Low[v], dfn[u]); }if(Dfn[v] = = Low[v]) {num++; Do{u = st[--top]; Vis[u] =false;        SCC[U] = num; } while(U! = v); }}BOOLWorkintMID) {init (); for(inti =0; I < mid; i++) {if(C[i] = =0) {Add_edge (A[i] + N, b[i]), Add_edge (B[i] + N, a[i]); }Else if(C[i] = =1) {Add_edge (A[i], b[i]), Add_edge (B[i], a[i]);        Add_edge (A[i] + N, B[i] + N), Add_edge (B[i] + N, A[i] + N); }Else if(C[i] = =2) {Add_edge (A[i], b[i] + N), Add_edge (B[i], a[i] + N); }    } for(inti =0; I <2*n; i++)if(Dfn[i] = =-1) Tarjan (i); for(inti =0; I < n; i++)if(Scc[i] = = Scc[i+n])return false;return true;}intMain () {intTscanf("%d", &t); while(t--) {scanf("%d%d", &n, &m); for(inti =0; I < m; i++)scanf("%d%d%d", &a[i], &b[i], &c[i]);intL =0, r = m, res; while(L <= R) {intMid = (L + r)/2;if(Work (mid)) L = mid +1, res = mid;ElseR = Mid-1; }printf("%d\n", res); }return 0;}