2017 ICPC网络赛(西安)--- Xor

题目连接

 

Problem

There is a tree with n nodes. For each node, there is an integer value ai, (1ai1,000,000,000 for 1in). There is q queries which are described as follow:

Assume the value on the path from node a to node b is t0,t1,tm. You are supposed to calculate t0 xor tk xor t2k xor ... xor tpk (pkm).

Input Format

There are multi datasets. (n50,000,q500,000).

For each dataset: In the first n1 lines, there are two integers u,v, indicates there is an edge connect node uand node v.

In the next nn lines, There is an integer ai (1ai1,000,000,000).

In the next q lines, There is three integers a,b and k. (1a,b,kn).

Output Format

For each query, output an integer in one line, without any additional space.

样例输入

5 6
1 5 4 1 2 1 3 2 19 26 0 8 17 5 5 1 1 3 2 3 2 1 5 4 2 3 4 4 1 4 5

样例输出

17
19 26 25 0 19

题意: 有一棵n个节点的树,每个节点上有个权值vi,现在q次询问:节点a到节点b的路径上,从a开始每k个节点跳一次所经过的所有节点的异或值为(a0,ak,a2k,a3k...)?

思路: 建树,倍增算法记录每个节点的深度和2^i的祖先,处理并记录每个节点i到根节点间隔为k(1,2,3……100)的异或值dp[i][k]。当k<=100时,使用记录的异或值dp计算a到b间隔为k的异或值;当k>100时,直接从a走到b,每次跳动使用倍增的信息(快速跳动)。

注:这道题是2017年打西安网络赛时没过的题,当时其实代码写的很接近了,测数据都没问题,一直检查不出来,我也一直惦记着这道题。本科毕业后读研,又看过一次还是没找出原因,今天五一放假,没啥事儿,我又看了一遍当时写的代码,突然发现没初始化fa数组,心想难道是计蒜客网站编译器不是默认未初始化的值为0?加上fa的初始化后提交了一把,过了!!!
My God! 心心念念的这道题竟然是这个原因没过,气呀。不过,今天总算是找到原因了,哈哈~ 又一次想起来西安正式赛的时候,一道铜牌题没过“LOL BP”导致没拿到银牌,可惜的是当时的银牌题都过了,唉~ 与银失之交臂。现在是研究生了,很少刷题了,以后要少看剧,多看看相关的图形学的专业书,充实自己,找个好工作。

代码如下:
  1 //https://nanti.jisuanke.com/t/A1273 《Xor》
  2 #include <iostream>
  3 #include <algorithm>
  4 #include <cstdio>
  5 #include <cstring>
  6 using namespace std;
  7 const int N = 5e4 + 5;
  8 int fa[N][20], deep[N], head[N];
  9 int v[N], cnt;
 10 bool vis[N];
 11 int dp[N][105];
 12 struct data
 13 {
 14     int to, next;
 15 }e[2 * N];
 16 
 17 void insert(int u, int v)
 18 {
 19     e[++cnt].to = v;
 20     e[cnt].next = head[u];
 21     head[u] = cnt;
 22     e[++cnt].to = u;
 23     e[cnt].next = head[v];
 24     head[v] = cnt;
 25 }
 26 int cal(int x, int t)
 27 {
 28     for (int i = 0; i <= 19; i++)
 29         if (t&(1 << i)) x = fa[x][i];
 30     return x;
 31 }
 32 void dfs(int x)
 33 {
 34     vis[x] = 1;
 35     for (int i = 1; i <= 19; i++)
 36     {
 37         if (deep[x]<(1 << i))break;
 38         fa[x][i] = fa[fa[x][i - 1]][i - 1];///倍增处理祖先信息
 39     }
 40     for (int k = 1; k <= 100; k++)
 41     {
 42         dp[x][k] = v[x];
 43         if (deep[x]<k) continue;
 44         int p = cal(x, k);
 45         dp[x][k] ^= dp[p][k];
 46     }
 47     for (int i = head[x]; i; i = e[i].next)
 48     {
 49         if (vis[e[i].to]) continue;
 50         deep[e[i].to] = deep[x] + 1;
 51         fa[e[i].to][0] = x;
 52         dfs(e[i].to);
 53     }
 54 }
 55 int lca(int x, int y)///求lca
 56 {
 57     if (deep[x]<deep[y]) swap(x, y);
 58     x = cal(x, deep[x] - deep[y]);
 59     for (int i = 19; i >= 0; i--)
 60         if (fa[x][i] != fa[y][i])
 61         {
 62             x = fa[x][i];
 63             y = fa[y][i];
 64         }
 65     if (x == y)return x;
 66     else return fa[x][0];
 67 }
 68 
 69 void init()
 70 {
 71     cnt = 0;
 72     memset(head, 0, sizeof(head));
 73     memset(vis, 0, sizeof(vis));
 74     memset(dp, 0, sizeof(dp));
 75     memset(deep, 0, sizeof(deep));
 76     memset(fa,0,sizeof(fa));
 77 }
 78 
 79 int main()
 80 {
 81     int n, q;
 82     while (scanf("%d%d", &n, &q) != EOF)
 83     {
 84         init();
 85         for (int i = 1; i<n; i++)
 86         {
 87             int x, y; scanf("%d%d", &x, &y);
 88             insert(x, y);
 89         }
 90         for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
 91         dfs(1);
 92         while (q--)
 93         {
 94             int x, y, k; scanf("%d%d%d", &x, &y, &k);
 95             int pa = lca(x, y);
 96             if (k <= 100)
 97             {
 98                 int ans = dp[x][k];
 99                 int h = (deep[x] - deep[pa]) % k;
100                 int t = k - h;
101                 if (deep[pa] >= t)
102                 {
103                     int l = cal(pa, t);
104                     ans ^= dp[l][k];
105                 }
106                 int r = k - h;
107                 t = deep[y] - deep[pa] - r;
108                 if (t<0) goto endw2;
109                 t %= k;
110                 y = cal(y, t);///
111                 ans ^= dp[y][k];
112                 t = k - r;
113                 if (deep[pa] >= t)
114                 {
115                     int l = cal(pa, t);
116                     ans ^= dp[l][k];
117                 }
118             endw2:;
119                 printf("%d\n", ans);
120             }
121             else
122             {
123                 int ans = 0;
124                 while (1)
125                 {
126                     ans ^= v[x];
127                     if (deep[x] - k<deep[pa]) break;
128                     x = cal(x, k);
129                 }
130                 int l = k - (deep[x] - deep[pa]);
131                 int t = deep[y] - deep[pa] - l;
132                 if (t<0) goto endw;
133                 t %= k;
134                 y = cal(y, t);
135                 while (1)
136                 {
137                     ans ^= v[y];
138                     if (deep[y] - k <= deep[pa]) break;
139                     y = cal(y, k);
140                 }
141             endw:;
142                 printf("%d\n", ans);
143             }
144         }
145     }
146     return 0;
147 }
148 /**
149 8 11
150 1 2
151 2 3
152 2 4
153 1 5
154 5 6
155 5 7
156 4 8
157 3 5 6 2 7 0 1 10
158 1 8 1
159 answer=14
160 */

 



posted @ 2019-05-01 17:27  茶飘香~  阅读(455)  评论(0编辑  收藏  举报