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

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

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 = ;
 int n, p;
 int ans[maxn], c[maxn*], 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 = ;
     while (x > ) {
         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, , sizeof(cur));
     while (!S.empty()) S.pop();
     S.push(u);
     len = ;
     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 ==  && p == ) return ;
         for (int i = ; i <= n; i++) g[i].clear();
         int x, y;
         for (int i = ; i < n; i++) {
             scanf("%d %d", &x, &y);
             g[x].push_back(y);
             g[y].push_back(x);
         }
         memset(c, , sizeof(c));
         dfs(p);
         for (int i = ; i <= n; i++) {
             ans[i] = sum(r[i]-) - sum(l[i]);
             add(l[i], );
         }
         for (int i = ; i <= n; i++)
             printf("%d%c", ans[i], i == n ? '\n' : ' ');
     }
     return ;
 }