题目传送门
1 /*
2 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^....)
3 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
4 关键在于更新异或和，精髓：a ^ b = c -> a ^ c = b, b ^ c = a;
5 */
6 #include <cstdio>
7 #include <cstring>
8 #include <cmath>
9 #include <algorithm>
10 #include <queue>
11 using namespace std;
12
13 const int MAXN = 7e4 + 10;
14 const int INF = 0x3f3f3f3f;
15 int ans[MAXN][2];
16 int d[MAXN], s[MAXN];
17
18 int main(void) //Codeforces Round #285 (Div. 2) C. Misha and Forest
19 {
20 // freopen ("C.in", "r", stdin);
21
22 int n;
23 while (scanf ("%d", &n) == 1)
24 {
25 queue<int> Q;
26 for (int i=0; i<n; ++i)
27 {
28 scanf ("%d%d", &d[i], &s[i]);
29 if (d[i] == 1) Q.push (i);
30 }
31
32 int cnt = 0;
33 while (!Q.empty ())
34 {
35 int u = Q.front (); Q.pop ();
36 if (d[u] == 0) continue;
37 int v = s[u];
38 ans[++cnt][0] = u; ans[cnt][1] = v;
39 if ((--d[v]) == 1) Q.push (v);
40 s[v] = s[v] ^ u;
41 }
42
43 printf ("%d
", cnt);
44 for (int i=1; i<=cnt; ++i)
45 {
46 printf ("%d %d
", ans[i][0], ans[i][1]);
47 }
48 }
49
50 return 0;
51 }