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Counting Offspring
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2424 Accepted Submission(s): 838
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
思路： 入栈前数目-入栈后数目.c++可调整栈的大小.#pragma comment(linker,"/STACK:100000000,100000000") 两个100000000均为字节数。
#pragma comment(linker,"/STACK:100000000,100000000") #include <cstdio> #include <string.h> using namespace std; const int MAXN=100005; struct Edge{ int to,net; }es[MAXN+MAXN]; int head[MAXN],tot; void addedge(int u,int v) { es[tot].to=v; es[tot].net=head[u]; head[u]=tot++; es[tot].to=u; es[tot].net=head[v]; head[v]=tot++; } int bit[MAXN]; int n,root; void add(int i,int x) { while(i<MAXN) { bit[i]+=x; i+=(i&(-i)); } } int sum(int i) { int s=0; while(i>0) { s+=bit[i]; i-=(i&(-i)); } return s; } int res[MAXN],tmp[MAXN]; void dfs(int u,int fa) { tmp[u]=sum(u-1); for(int i=head[u];i!=-1;i=es[i].net) { int v=es[i].to; if(v!=fa) { add(v,1); dfs(v,u); } } res[u]=sum(u-1)-tmp[u]; } int main() { while(scanf("%d%d",&n,&root)!=EOF&&(n||root)) { memset(head,-1,sizeof(head)); memset(bit,0,sizeof(bit)); tot=0; for(int i=0;i<n-1;i++) { int u,v; scanf("%d%d",&u,&v); addedge(u,v); } dfs(root,-1); for(int i=1;i<n;i++) { printf("%d ",res[i]); } printf("%d\n",res[n]); } return 0; }
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原文地址：http://www.cnblogs.com/program-ccc/p/5693006.html
