Hdu 3887树状数组+模拟栈

题目链接

Counting Offspring

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1757    Accepted Submission(s): 582

Problem Description

You are given a tree, it’s root is p, and the node is numbered from 1 to n. Now define f(i) as the number of nodes whose number is less than i in all the succeeding nodes of node i. Now we need to calculate f(i) for any possible i.

Input

Multiple cases (no more than 10), for each case:
The first line contains two integers n (0<n<=10^5) and p, representing this tree has n nodes, its root is p.
Following n-1 lines, each line has two integers, representing an edge in this tree.
The input terminates with two zeros.

Output

For each test case, output n integer in one line representing f(1), f(2) … f(n), separated by a space.

Sample Input

15 7 7 10 7 1 7 9 7 3 7 4 10 14 14 2 14 13 9 11 9 6 6 5 6 8 3 15 3 12 0 0

Sample Output

0 0 0 0 0 1 6 0 3 1 0 0 0 2 0

可以直接dfs（需要手工加栈）

Accepted Code:

/*************************************************************************
    > File Name: 3887.cpp
    > Author: Stomach_ache
    > Mail: sudaweitong@gmail.com
    > Created Time: 2014年08月09日 星期六 14时11分33秒
    > Propose: 
 ************************************************************************/
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cmath>
#include <string>
#include <cstdio>
#include <vector>
#include <fstream>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100002;
int n, p;
int ans[maxn], c[maxn], tmp[maxn];
bool vis[maxn];
vector<int> g[maxn];

int lowbit(int x) {
      return x & -x;
}

int sum(int x) {
      int res = 0;
    while (x > 0) {
          res += c[x];
        x -= lowbit(x);
    }
    return res;
}

void add(int x, int v) {
      while (x <= n) {
          c[x] += v;
        x += lowbit(x);
    }
}

void dfs(int u, int fa) {
      tmp[u] = sum(u-1);
    for (int i = 0; i < (int)g[u].size(); i++) {
          int v = g[u][i];
          if (v == fa) continue;
          add(v, 1);
          dfs(v, u);
    }    
    ans[u] = sum(u-1) - tmp[u];
}

int main(void) {
      while (~scanf("%d %d", &n, &p)) {
          if (n == 0 && p == 0) return 0;
          for (int i = 1; i <= n; i++) g[i].clear();
        int x, y;
        for (int i = 1; i < n; i++) {
            scanf("%d %d", &x, &y);
            g[x].push_back(y);
            g[y].push_back(x);
        }
        memset(c, 0, sizeof(c));
        dfs(p, -1);
        for (int i = 1; i <= n; i++) 
              printf("%d%c", ans[i], i == n ? '
' : ' ');
    }
    return 0;
}

附上模拟栈的AC代码：

/*************************************************************************
    > File Name: 3887.cpp
    > Author: Stomach_ache
    > Mail: sudaweitong@gmail.com
    > Created Time: 2014年08月09日 星期六 14时11分33秒
    > Propose: 
 ************************************************************************/
#include <stack>
#include <cmath>
#include <string>
#include <cstdio>
#include <vector>
#include <fstream>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100002;
int n, p;
int ans[maxn], c[maxn*2], l[maxn], r[maxn], cur[maxn];
int len;
bool vis[maxn];
vector<int> g[maxn];
stack<int> S;

int lowbit(int x) {
      return x & -x;
}

int sum(int x) {
    int res = 0;
    while (x > 0) {
        res += c[x];
        x -= lowbit(x);
    }
    return res;
}

void add(int x, int v) {
    while (x <= len) {
        c[x] += v;
        x += lowbit(x);
    }
}

void dfs(int u) {
    memset(vis, false, sizeof(vis));
    memset(cur, 0, sizeof(cur));
    while (!S.empty()) S.pop();
    S.push(u);
    len = 0;
    while (!S.empty()) {
        int now = S.top();
        if (!vis[now]) {
            vis[now] = true;
            l[now] = ++len;
        }
        bool flag = false;
        for (int& i = cur[now]; i < (int)g[now].size(); i++) {
              int v = g[now][i];
              if (!vis[v]) {
                S.push(v);
                flag = true;
                break;
            }
        }
        if (flag) continue;
        if (vis[now]) {
            r[now] = ++len;
            S.pop();
        }
    }
}

int main(void) {
      while (~scanf("%d %d", &n, &p)) {
        if (n == 0 && p == 0) return 0;
        for (int i = 1; i <= n; i++) g[i].clear();
        int x, y;
        for (int i = 1; i < n; i++) {
            scanf("%d %d", &x, &y);
            g[x].push_back(y);
            g[y].push_back(x);
        }
        memset(c, 0, sizeof(c));
        dfs(p);
        for (int i = 1; i <= n; i++) {
            ans[i] = sum(r[i]-1) - sum(l[i]);
            add(l[i], 1);
        }
        for (int i = 1; i <= n; i++) 
            printf("%d%c", ans[i], i == n ? '
' : ' ');
    }
    return 0;
}