"HDU" 1814 peaceful Commission

http://acm.hdu.edu.cn/showproblem.php?pid=1814

Test instructions: N 2-person group, numbered 2n and 2n+1 respectively, each group selected a person to come out, and give the M-bar relationship (x, y) to select X can not choose Y, ask whether can choose 1 people from each group. And the smallest answer to the output dictionary order. (n<=8000, m<=20000)

#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include < iostream>using namespace Std;const int n=8005, m=20005;struct E {int next, to;} E[m<<1];int CNT, ihead[n<<1], top, s[n<<1], N, M;bool vis[n<<1];void Add (int u, int v) {e[++cnt]= E {Ihead[u], v}; ihead[u]=cnt; }bool dfs (int x) {if (vis[x^1]) return 0;if (Vis[x]) return 1;vis[x]=1;s[++top]=x;for (int i=ihead[x]; i; i=e[i].next) if (!d FS (e[i].to)) return 0;return 1;} BOOL Work () {int nn=n<<1;for (int i=0, i<nn; i+=2) if (!vis[i] &&!vis[i+1]) {top=0;if (!dfs (i)) {while (top ) vis[s[top--]]=0;if (!dfs (i+1)) return 0;} for (int i=0; i<nn; ++i) if (Vis[i]) printf ("%d\n", i+1); return 1;} int main () {while (~scanf ("%d%d", &n, &m)) {for (int i=0; i<m; ++i) {int x, y; scanf ("%d%d", &x, &y);-- X --y; Add (x, y^1); Add (y, x^1); }if (!work ()) puts ("NIE"), Cnt=top=0;memset (Vis, 0, sizeof (BOOL) * (n<<1)), memset (ihead, 0, sizeof (int) * (n<<1));} return 0;}

　　

The problem is obvious 2-sat question = =

For relationships (x, y) is actually satisfying! (X & Y)

That is, when the x=1 y must be 0, that is, the Edge x->y '

When Y=1 x must be 1, that is, the Edge y->x '

Because the DFS algorithm itself is the dictionary order of the smallest = = directly engage in the line = =

(MOM, this algorithm is O (nm) Well qaq seems to have to write tarjan later = = (although the dictionary order minimum can only use this O (nm) algorithm = =? ）

"HDU" 1814 peaceful Commission