水题 Codeforces Round #285 (Div. 2) C. Misha and Forest

题目传送门

 /*
     题意：给出无向无环图，每一个点的度数和相邻点的异或和(a^b^c^....)
     图论/位运算：其实这题很简单。类似拓扑排序，先把度数为1的先入对，每一次少一个度数
                 关键在于更新异或和，精髓：a ^ b = c -> a ^ c = b, b ^ c = a;
 */
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <queue>
 using namespace std;

 const int MAXN = 7e4 + ;
 const int INF = 0x3f3f3f3f;
 int ans[MAXN][];
 int d[MAXN], s[MAXN];

 int main(void)        //Codeforces Round #285 (Div. 2) C. Misha and Forest
 {
     // freopen ("C.in", "r", stdin);

     int n;
     while (scanf ("%d", &n) == )
     {
         queue<int> Q;
         for (int i=; i<n; ++i)
         {
             scanf ("%d%d", &d[i], &s[i]);
             if (d[i] == )    Q.push (i);
         }

         int cnt = ;
         while (!Q.empty ())
         {
             int u = Q.front ();    Q.pop ();
             if (d[u] == )    continue;
             int v = s[u];
             ans[++cnt][] = u;    ans[cnt][] = v;
             if ((--d[v]) == )    Q.push (v);
             s[v] = s[v] ^ u;
         }

         printf ("%d\n", cnt);
         for (int i=; i<=cnt; ++i)
         {
             printf ("%d %d\n", ans[i][], ans[i][]);
         }
     }

     return ;
 }