标签：des style blog http color java os io
Counting Offspring
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1757 Accepted Submission(s): 582Problem DescriptionYou 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.
InputMultiple 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.
OutputFor each test case, output n integer in one line representing f(1), f(2) … f(n), separated by a space.
Sample Input15 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 Output0 0 0 0 0 1 6 0 3 1 0 0 0 2 0
1 /*************************************************************************
2 > File Name: 3887.cpp
3 > Author: Stomach_ache
4 > Mail: sudaweitong@gmail.com
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: sudaweitong@gmail.com
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 }
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标签：des style blog http color java os io
原文地址：http://www.cnblogs.com/Stomach-ache/p/3901652.html
