POJ 2492 (and check set)

Test instructions: One scientist put forward the hypothesis that a worm is heterosexual, not homosexual, and then starts experimenting to verify conjecture, gives n bugs, numbers from 1 to N, gives the number of the Q Group of love worms, and asks whether to verify conjecture.
Puzzle: Array multiplication, give each worm a heterosexual object collection, and then each time the input of the two worms if not within the same set, placed in the other heterosexual set, or can not verify the conjecture.

#include <stdio.h>ConstintN =4005;intNQ, Pa[n];intGet_parent (int x) {return x= = pa[x] ?x: pa[x] = Get_parent (pa[x]);}intMain () {intT, CAS =1; scanf"%d", &t); while(t--) {scanf ("%d%d", &n, &Q);intnn =2N for(inti =1; I <= nn; i++) Pa[i] = i;intFlag =0; while(Q--) {intA, B; scanf"%d%d", &a, &b);intpx = Get_parent (a);intPY = get_parent (b);if(px = = PY) flag =1;Else{intT1 = get_parent (a + N);intt2 = get_parent (b + N);                PA[PX] = T2;            Pa[py] = T1; }        }if(CAs >1)printf("\ n");printf("Scenario #%d: \ n", cas++);if(flag)printf("Suspicious bugs found!\n");Else            printf("No suspicious bugs found!\n"); }return 0;}

Another approach, with the right and check set, plus an array rel[i] indicates whether I and parent node is same-sex, looking for root node update rel[i], if found that the input of the two worms is the same root node and the same root node relationship, the two insects are homosexual, otherwise put in a set, and update the rel of the root node of one of the worms.

#include <stdio.h>ConstintN =4005;intNQ, Pa[n], rel[n];intGet_parent (int x,int&r) {inttemp =x; R =1; while(x! = pa[x]) {if(rel[x] ==0) R =!r;x= pa[x]; } Pa[temp] =x; Rel[temp] = r;returnpa[x];}intMain () {intT, CAS =1; scanf"%d", &t); while(t--) {scanf ("%d%d", &n, &Q); for(inti =1; I <= N;            i++) {Pa[i] = i; Rel[i] =1; }intFlag =0; while(Q--) {intA, B, R1, R2; scanf"%d%d", &a, &b);intpx = Get_parent (A, R1);intPY = Get_parent (b, R2);if(px = = Py && R1 = = r2) flag =1;Else{Pa[py] = px; Rel[py] = R1 = = r2?0:1; }        }if(CAs >1)printf("\ n");printf("Scenario #%d: \ n", cas++);if(flag)printf("Suspicious bugs found!\n");Else            printf("No suspicious bugs found!\n"); }return 0;}

POJ 2492 (and check set)