uva1572--self-set (World Final 2013, graph theory model)

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This question is worth learning:
1. Enter the matching code very beautifully!
2. Map the a+,a-into 2n,2n+1 and then reverse it by (2n) ^1 = 2n+1, (2n+1) ^1 = 2n, and have to say it is very subtle!

The puzzle: Since the square is infinite, the link number and other extensible numbers on the linked square as nodes, the linked square as the edge, looking for the secondary graph of the ring.
Examples in the topic:
A-|->->->->->->->->->->a+|->->->->->->->-> (A+,A+,00)
The node to connect ———— two square connection points ———— the other nodes of the square being linked

#include <cstdio>#include <cstring>intg[ -][ -];intvis[ -];//Turn a+-into 2n and 2n+1intID (CharXChary);//This function indicates that an interface is connected to the other interfaces that are converted into the U//Use (2n) ^1 = 2n+1, (2n+1) ^1 = 2n to invertvoidContCharX1,CharX2,CharY1,Chary2);BOOLDfsintu);BOOLCyc ();intMain () {intN while(scanf("%d", &n) = =1&& N) {memsetG' + ',sizeof(G));memset(Vis,0,sizeof(VIS)); for(inti =0; I < n; i++) {Chars[9];scanf('%s ', s); for(intj =0; J <4; J + +) for(intK =0; K <4; k++)if(j! = k)//For each edge connected to other non-self edgesCont (s[j*2],s[j*2+1],s[k*2],s[k*2+1]); }if(Cyc ())puts("unbounded");//This is the infinite ...         Else            puts("bounded"); }return 0;}intID (CharXCharY) {return(X-' A ')*2+ (y = =' + '?0:1);}voidContCharX1,CharX2,CharY1,Chary2) {if(X1 = =' 0 '|| Y1 = =' 0 ')return;intU = ID (x1,x2) ^1;intv = ID (y1,y2); g[u][v]++;}BOOLDfsintu) {Vis[u] =-1; for(inti =0; I < the; i++) {if(G[u][i]) {if(Vis[i] <0)return true;Else if(!vis[i] && dfs (i))return true; }} Vis[u] =1;return false;}BOOLCyc () { for(inti =0; I < the; i++)if(!vis[i] && dfs (i))return true;return false;}

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uva1572--self-set (World Final 2013, graph theory model)