HDU 6161 （2017多校9

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题意：

给你一个n （n <= 1e8）个节点的完全二叉树，开始时，每个节点的权值为他本身，两种操作：

1. 查询经过节点x 的最大权值链。

2. 改变x 的权值为v

思路：

n是1e8，没法建树。

但他是一个完全二叉树，一条链就log n 个点。

那么我们每次修改权值时，维护一下这条链的信息。那么我们只需要n log n 的空间就足够了。

所有我们令dp[x]表示从x 往下走最大权值是多少。

val[x]表示 x 这个点权值是多少，如果没有，表示为本身（初始状态）

然后就是模拟就行了。

#include <cstdio>#include <cstring>#include <algorithm>#include <map>using namespace std;map<int, long long >dp, val;char s[10];int n, q;long long calc(int x){    if (x > n) return 0;    if (dp.count(x)) return dp[x];    int d1 = 0;    int d2 = 0;    int x2 = x;    while(x2 <= n){        ++d1;        x2 = x2 * 2 + 1;    }    int x3 = x;    while(x3 <= n){        ++d2;        x3 = x3 * 2;    }    long long sum = 0;    if (x2 > n){        x2 >>= 1;    }    while(x2 != x){        sum += x2;        x2 /= 2;    }    sum += x2;    long long ans = sum;    sum = 0;    if (d1 != d2){        x2 = n;        while(x2 != x){            sum += x2;            x2 /= 2;        }        sum += x2;        ans = max(ans, sum);    }    return ans;}void change(int x,long long v){    val[x] = v;    dp[x] = max(calc(x<<1), calc(x << 1 | 1)) + v;    int md = x % 2;    long long ans = max(calc(x<<1) , calc(x << 1 | 1)) + v;    x/= 2;    while(x > 1){        long long t;        if (val.count(x)) t = val[x];        else t = x;//        ans += t;        if (md){            dp[x] = max(calc(x << 1), ans) + t;        }        else {            dp[x] = max(ans, calc(x<<1|1) ) + t;        }        ans = dp[x];        md = x % 2;        x /= 2;    }}long long query(int x){    long long t;    if (val.count(x)) t = val[x];    else t = x;    long long ans = calc(x << 1) + calc(x << 1 | 1) + t;    long long tmp = calc(x);    int md = x % 2;    x >>= 1;    while(x >= 1){        if (val.count(x))t = val[x];        else t = x;        tmp += t;        if (md){            ans = max(ans, calc(x << 1) + tmp);        }        else {            ans = max(ans, calc(x<<1|1) + tmp);        }        md = x % 2;        x /= 2;    }    return ans;}int main(){    while(~scanf("%d %d",&n, &q)){        val.clear();        dp.clear();        for (int i = 0; i < q; ++i){            scanf("%s", s);            if (s[0] == 'q'){                int x;                scanf("%d",&x);                long long ans = query(x);                printf("%I64d\n", ans);            }            else {                int x;                long long v;                scanf("%d%I64d", &x, &v);                change(x, v);            }        }    }    return 0;}

Big binary tree

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 492 Accepted Submission(s): 168

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

2017 Multi-University Training Contest - Team 9

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