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BZOJ2819 Nim

时间:2014-12-14 00:31:57      阅读:311      评论:0      收藏:0      [点我收藏+]

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做法。。。就不讲了,参见hzwer的blog好了

我们发现只要维护树上点到根的xor值就可以了,于是先搞个dfs序,然后用树状数组维护即可。

反正各种调不出。。。各种WA

后来发现又是LCA的姿势不对= =,今天不是刚写过noip题嘛T T

蒟蒻还是滚去挖矿算了、、、

 

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  1 /**************************************************************
  2     Problem: 2819
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:10240 ms
  7     Memory:76648 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <algorithm>
 12  
 13 #define lowbit(x) x & -x
 14 using namespace std;
 15 const int N = 500005;
 16  
 17 struct edge {
 18     int next, to;
 19     edge() {}
 20     edge(int _n, int _t) : next(_n), to(_t) {}
 21 } e[N << 1];
 22  
 23 struct tree_node {
 24     int v, dep, st, ed;
 25     int fa[19];
 26 } tr[N];
 27  
 28 int n;
 29 int tot, first[N];
 30 int cnt_seq, BIT[N];
 31  
 32 inline int read() {
 33     int x = 0;
 34     char ch = getchar();
 35     while (ch < 0 || 9 < ch)
 36         ch = getchar();
 37     while (0 <= ch && ch <= 9) {
 38         x = x * 10 + ch - 0;
 39         ch = getchar();
 40     }
 41     return x;
 42 }
 43  
 44 void XOR(int x, int v) {
 45     while (x <= n)
 46         BIT[x] ^= v, x += lowbit(x);
 47 }
 48  
 49 int query(int x) {
 50     int res = 0;
 51     while (x)
 52         res ^= BIT[x], x -= lowbit(x);
 53     return res;
 54 }
 55  
 56 inline void Add_Edges(int x, int y) {
 57     e[++tot] = edge(first[x], y), first[x] = tot;
 58     e[++tot] = edge(first[y], x), first[y] = tot;
 59 }
 60  
 61 void dfs(int p) {
 62     int x, y;
 63     for (x = 1; x <= 18; ++x)
 64         tr[p].fa[x] = tr[tr[p].fa[x - 1]].fa[x - 1];
 65     tr[p].st = ++cnt_seq;
 66     for (x = first[p]; x; x = e[x].next)
 67         if ((y = e[x].to) != tr[p].fa[0]) {
 68             tr[y].fa[0] = p, tr[y].dep = tr[p].dep + 1;
 69             dfs(y);
 70         }
 71     tr[p].ed = cnt_seq;
 72     XOR(tr[p].st, tr[p].v), XOR(tr[p].ed + 1, tr[p].v);
 73 }
 74  
 75 int lca(int x, int y) {
 76     int i;
 77     if (tr[x].dep < tr[y].dep) swap(x, y);
 78     for (i = 18; ~i; --i)
 79         if (tr[tr[x].fa[i]].dep >= tr[y].dep)
 80             x = tr[x].fa[i];
 81     if (x == y) return x;
 82     for (i = 18; ~i; --i)
 83         if (tr[x].fa[i] != tr[y].fa[i])
 84             x = tr[x].fa[i], y = tr[y].fa[i];
 85     return tr[x].fa[0];
 86 }
 87  
 88 int main() {
 89     int i, x, y, LCA, Q;
 90     char ch;
 91     n = read();
 92     for (i = 1; i <= n; ++i)
 93         tr[i].v = read();
 94     for (i = 1; i < n; ++i)
 95         Add_Edges(read(), read());
 96     tr[1].dep = 1;
 97     dfs(1);
 98     Q = read();
 99     while (Q--) {
100         ch = getchar();
101         while (ch != Q && ch != C) ch = getchar();
102         x = read(), y = read();
103         if (ch == Q)
104             puts(query(tr[x].st) ^ query(tr[y].st) ^ tr[lca(x, y)].v ? "Yes" : "No");
105         else {
106             XOR(tr[x].st, tr[x].v), XOR(tr[x].ed + 1, tr[x].v);
107             tr[x].v = y;
108             XOR(tr[x].st, tr[x].v), XOR(tr[x].ed + 1, tr[x].v);
109         }
110     }
111     return 0;
112 }
View Code

 

BZOJ2819 Nim

标签:style   blog   http   io   ar   color   os   sp   for   

原文地址:http://www.cnblogs.com/rausen/p/4162079.html

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