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:

 1 /*************************************************************************  2  > File Name: 3887.cpp  3  > Author: Stomach_ache  4  > Mail: [email protected]  5  > Created Time: 2014年08月09日 星期六 14时11分33秒  6  > Propose:  7  ************************************************************************/

 8 #pragma comment(linker, "/STACK:1024000000,1024000000")

 9 #include <cmath>

10 #include <string>

11 #include <cstdio>

12 #include <vector>

13 #include <fstream>

14 #include <cstring>

15 #include <iostream>

16 #include <algorithm>

17 using namespace std; 18 const int maxn = 100002; 19 int n, p; 20 int ans[maxn], c[maxn], tmp[maxn]; 21 bool vis[maxn]; 22 vector<int> g[maxn]; 23 

24 int lowbit(int x) { 25       return x & -x; 26 } 27 

28 int sum(int x) { 29       int res = 0; 30     while (x > 0) { 31           res += c[x]; 32         x -= lowbit(x); 33  } 34     return res; 35 } 36 

37 void add(int x, int v) { 38       while (x <= n) { 39           c[x] += v; 40         x += lowbit(x); 41  } 42 } 43 

44 void dfs(int u, int fa) { 45       tmp[u] = sum(u-1); 46     for (int i = 0; i < (int)g[u].size(); i++) { 47           int v = g[u][i]; 48           if (v == fa) continue; 49           add(v, 1); 50  dfs(v, u); 51  } 52     ans[u] = sum(u-1) - tmp[u]; 53 } 54 

55 int main(void) { 56       while (~scanf("%d %d", &n, &p)) { 57           if (n == 0 && p == 0) return 0; 58           for (int i = 1; i <= n; i++) g[i].clear(); 59         int x, y; 60         for (int i = 1; i < n; i++) { 61             scanf("%d %d", &x, &y); 62  g[x].push_back(y); 63  g[y].push_back(x); 64  } 65         memset(c, 0, sizeof(c)); 66         dfs(p, -1); 67         for (int i = 1; i <= n; i++) 68               printf("%d%c", ans[i], i == n ? '\n' : ' '); 69  } 70     return 0; 71 }

附上模拟栈的AC代码：

 1 /*************************************************************************  2  > File Name: 3887.cpp  3  > Author: Stomach_ache  4  > Mail: [email protected]  5  > Created Time: 2014年08月09日 星期六 14时11分33秒  6  > Propose:  7  ************************************************************************/

 8 #include <stack>

 9 #include <cmath>

10 #include <string>

11 #include <cstdio>

12 #include <vector>

13 #include <fstream>

14 #include <cstring>

15 #include <iostream>

16 #include <algorithm>

17 using namespace std; 18 const int maxn = 100002; 19 int n, p; 20 int ans[maxn], c[maxn*2], l[maxn], r[maxn], cur[maxn]; 21 int len; 22 bool vis[maxn]; 23 vector<int> g[maxn]; 24 stack<int> S; 25 

26 int lowbit(int x) { 27       return x & -x; 28 } 29 

30 int sum(int x) { 31     int res = 0; 32     while (x > 0) { 33         res += c[x]; 34         x -= lowbit(x); 35  } 36     return res; 37 } 38 

39 void add(int x, int v) { 40     while (x <= len) { 41         c[x] += v; 42         x += lowbit(x); 43  } 44 } 45 

46 void dfs(int u) { 47     memset(vis, false, sizeof(vis)); 48     memset(cur, 0, sizeof(cur)); 49     while (!S.empty()) S.pop(); 50  S.push(u); 51     len = 0; 52     while (!S.empty()) { 53         int now = S.top(); 54         if (!vis[now]) { 55             vis[now] = true; 56             l[now] = ++len; 57  } 58         bool flag = false; 59         for (int& i = cur[now]; i < (int)g[now].size(); i++) { 60               int v = g[now][i]; 61               if (!vis[v]) { 62  S.push(v); 63                 flag = true; 64                 break; 65  } 66  } 67         if (flag) continue; 68         if (vis[now]) { 69             r[now] = ++len; 70  S.pop(); 71  } 72  } 73 } 74 

75 int main(void) { 76       while (~scanf("%d %d", &n, &p)) { 77         if (n == 0 && p == 0) return 0; 78         for (int i = 1; i <= n; i++) g[i].clear(); 79         int x, y; 80         for (int i = 1; i < n; i++) { 81             scanf("%d %d", &x, &y); 82  g[x].push_back(y); 83  g[y].push_back(x); 84  } 85         memset(c, 0, sizeof(c)); 86  dfs(p); 87         for (int i = 1; i <= n; i++) { 88             ans[i] = sum(r[i]-1) - sum(l[i]); 89             add(l[i], 1); 90  } 91         for (int i = 1; i <= n; i++) 92             printf("%d%c", ans[i], i == n ? '\n' : ' '); 93  } 94     return 0; 95 }