Big binary tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 597    Accepted Submission(s): 207

Problem Description

You are given a complete binary tree with n nodes. The root node is numbered 1, and node x's father node is

⌊x/2⌋. At the beginning, node x has a value of exactly x. We define the value of a path as the sum of all nodes it passes(including two ends, or one if the path only has one node). Now there are two kinds of operations:

1.  change u x Set node u's value as x(1≤u≤n;1≤x≤10^10)

2.  query u Query the max value of all paths which passes node u.

 

 

Input

There are multiple cases.

For each case:

The first line contains two integers n,m(1≤n≤10^8,1≤m≤10^5), which represent the size of the tree and the number of operations, respectively.

Then m lines follows. Each line is an operation with syntax described above.

 

 

Output

For each query operation, output an integer in one line, indicating the max value of all paths which passes the specific node.

 

 

Sample Input

6 13query 1query 2query 3query 4query 5query 6change 6 1query 1query 2query 3query 4query 5query 6

 

Sample Output

171717161717121212111212

 

Source

题目链接：

分析：考虑dp，f(x)表示从点x开始向下走得到的最大的点权和，查询直接从x开始向上走更新答案即可。

考虑快速算 f(x) 对于子树内没有被修改过的点的 f(x) 可以快速分类讨论算出，而不满足本条件的点只有 O(mlogm) 个，在hash上dp即可。

下面给出AC代码：

1 #include
      
        2 using namespace std; 3 #define pb push_back 4 #define mkp make_pair 5 #define fi first 6 #define se second 7 #define ll long long 8 #define M 1000000007 9 #define all(a) a.begin(), a.end()10 11 int n, m;12 char s[20];13 map
       
         mp;14 map
        
          amp;15 16 inline ll askmax(int u){17     if(u > n) return 0;18     if(mp.count(u)) return mp[u];19     else{20         int l = u, r = u;21         while(l * 2 <= n){22             l <<= 1;23             r = (r << 1) | 1;24         }25         r = min(r, n);26         ll res = 0;27         while(r >= u) res += r, r >>= 1;28         return res;29     }30 }31 32 inline int ask(int x){33     return amp.count(x) ? amp[x] : x;34 }35 36 int main(){37     while(~scanf("%d%d", &n, &m)){38         mp.clear();39         amp.clear();40         while(m--){41             int x, v;42             scanf("%s", s);43             if(s[0] == 'c'){44                 scanf("%d%d", &x, &v);45                 amp[x] = v;46                 for(; x; x >>= 1)47                     mp[x] = max(askmax(x << 1), askmax((x << 1) | 1)) + ask(x);48             }else{49                 scanf("%d", &x);50                 int px = x;51                 ll res = 0, now = 0;52                 for(; x >> 1;){53                     bool k = ~x & 1; x >>= 1;54                     now += ask(x);55                     ll tmp = askmax(x << 1 | k);56                     if(now + tmp > res) res = now + tmp;57                 }58                 res += askmax(px);59                 res = max(res, askmax(px << 1) + askmax(px << 1 | 1) + ask(px));60                 printf("%lld\n", res);61             }62         }63     }64     65 #ifndef ONLINE_JUDGE66     system("pause");67 #endif68     return 0;69 }