图论/位运算 Codeforces Round #285 (Div. 2) C. Misha and Forest

 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 }
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