HDU 3887 Counting Offspring

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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 77 107 17 97 37 410 1414 214 139 119 66 56 83 153 120 0

Sample Output

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

dfs序+树状数组+数组模拟链表来存储树

# include<stdio.h># include<math.h># include<string.h>#pragma comment(linker, "/STACK:1024000000,1024000000")  //防爆栈struct abc{    int next,just;};abc f[300000];int n, e, root;int frist[100005], c[100005], d[100005];int add(int x, int y){    f[e].just = y; f[e].next = frist[x]; frist[x] = e++;    f[e].just = x; f[e].next = frist[y]; frist[y] = e++;    return 0;}int lowb(int t){ return t&(-t); }int fadd(int x){    while (x <= n) { d[x]++; x += lowb(x); }    return 0;}int sum(int x){    if (x <= 0) return 0;    return d[x] + sum(x - lowb(x));}int ss(int x, int fa){    int a = sum(x);    int k = frist[x];    while (k > 0)    {        if (f[k].just != fa)        {            fadd(f[k].just);            ss(f[k].just, x);        }        k = f[k].next;    }    c[x] = sum(x) - a;    return 0;}int main(){    while (scanf("%d%d", &n, &root) != EOF&&n+root)    {        memset(f,0,sizeof(f));        memset(frist, 0, sizeof(frist));        memset(d, 0, sizeof(d));        e = 1;        for (int i = 1; i < n; i++)        {            int a, b;            scanf("%d%d", &a, &b);            add(a, b);        }        ss(root,root);        printf("%d", c[1]);        for (int i =2; i <= n; i++) printf(" %d", c[i]);        printf("\n");    }    return 0;}

hdu上这个已经过不了了，只能手动模拟栈。

# include<stdio.h># include<math.h># include<string.h> # include<stack># include<algorithm>using namespace std;stack<int> p;struct wrz{int next, just;};wrz f[300000];int n, e, root;int frist[100005], c[100005], d[100005], u[100005];int add(int x, int y){f[e].just = y; f[e].next = frist[x]; frist[x] = e++;f[e].just = x; f[e].next = frist[y]; frist[y] = e++;return 0;}int lowb(int t){ return t&(-t); }int fadd(int x){while (x <= n) { d[x]++; x += lowb(x); }return 0;}int sum(int x){if (x <= 0) return 0;return d[x] + sum(x - lowb(x));}int ss(int x){p.push(x);u[x] = 1;while (!p.empty()){begin:int x = p.top();if (c[x] == 0) {fadd(x);c[x] = sum(x);}int k = frist[p.top()];while (k > 0){if (u[f[k].just] == 0){u[f[k].just] = 1;p.push(f[k].just);goto begin;}k = f[k].next;}c[x] = sum(x) - c[x];p.pop();}return 0;}int main(){while (~scanf("%d%d", &n, &root), n + root){memset(f, 0, sizeof(f));memset(frist, 0, sizeof(frist));memset(d, 0, sizeof(d));memset(u, 0, sizeof(u));memset(c, 0, sizeof(c));e = 1;for (int i = 1; i < n; i++){int a, b;scanf("%d%d", &a, &b);add(a, b);}ss(root);printf("%d", c[1]);for (int i = 2; i <= n; i++) printf(" %d", c[i]);printf("\n");}return 0;}

上面这个是用了goto的，下面的是普通的。

# include<stdio.h># include<math.h># include<string.h> # include<stack># include<algorithm>using namespace std;stack<int> p;struct wrz{int next, just;};wrz f[300000];int n, e, root;int frist[100005], c[100005], d[100005], u[100005];int add(int x, int y){f[e].just = y; f[e].next = frist[x]; frist[x] = e++;f[e].just = x; f[e].next = frist[y]; frist[y] = e++;return 0;}int lowb(int t){ return t&(-t); }int fadd(int x){while (x <= n) { d[x]++; x += lowb(x); }return 0;}int sum(int x){if (x <= 0) return 0;return d[x] + sum(x - lowb(x));}int ss(int x){p.push(x);u[x] = 1;while (!p.empty()){int x = p.top();if (c[x] == 0) { fadd(x); c[x] = sum(x); }for (int k = frist[x]; k; k = f[k].next)if (u[f[k].just] == 0){u[f[k].just] = 1;p.push(f[k].just);}if (p.top() == x){c[x] = sum(x) - c[x];p.pop();}}return 0;}int main(){while (~scanf("%d%d", &n, &root), n + root){memset(f, 0, sizeof(f));memset(frist, 0, sizeof(frist));memset(d, 0, sizeof(d));memset(u, 0, sizeof(u));memset(c, 0, sizeof(c));e = 1;for (int i = 1; i < n; i++){int a, b;scanf("%d%d", &a, &b);add(a, b);}ss(root);printf("%d", c[1]);for (int i = 2; i <= n; i++) printf(" %d", c[i]);printf("\n");}return 0;}