HDU 3887 Counting Offspring(dfs序的应用)

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题意：给你一棵树，每个节点有个标号，求树上每个节点其子树中有多少个节点标号比自己小。

做法我想了半天，知道可以用树状数组统计，但是如果删除影响想了好久。。看了题解才知道原来怎么简单，根本不需要删除其他点的影响，直接在遍历自己下面子树之前统计一下，遍历后再统计一下，相减就是子树里的。脑子还是太榆木了。

另外因为树比较大，不扩栈的话可以用栈模拟。

AC代码：

#pragma comment(linker, "/STACK:102400000,102400000")#include<cstdio>#include<ctype.h>#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdlib>#include<stack>#include<queue>#include<set>#include<map>#include<cmath>#include<ctime>#include<string.h>#include<string>#include<sstream>#include<bitset>using namespace std;#define ll long long#define ull unsigned long long#define eps 1e-8#define NMAX 201000#define MOD 1000000#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define PI acos(-1)template<class T>inline void scan_d(T &ret){    char c;    int flag = 0;    ret=0;    while(((c=getchar())<'0'||c>'9')&&c!='-');    if(c == '-')    {        flag = 1;        c = getchar();    }    while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();    if(flag) ret = -ret;}const int maxn = 100100;struct Edge{    int v,next;}e[maxn*2];int head[maxn];int nct,n;int T[maxn],dp[maxn];void add_edge(int u, int v){    e[nct].v = v;    e[nct].next = head[u]; head[u] = nct++;    e[nct].v = u;    e[nct].next = head[v]; head[v] = nct++;}inline int lowbit(int x){    return x&(-x);}void add(int x, int d){    while(x <= n)    {        T[x] += d;        x += lowbit(x);    }}int sum(int x){    int ret = 0;    while(x > 0)    {        ret += T[x];        x -= lowbit(x);    }    return ret;}void dfs(int u, int fa){    int t1 = sum(u-1);    for(int i = head[u]; i != -1; i = e[i].next)    {        int v = e[i].v;        if(v == fa) continue;        dfs(v,u);    }    dp[u] = sum(u-1)-t1;    add(u,1);}int main(){#ifdef GLQ    freopen("input.txt","r",stdin);//    freopen("o2.txt","w",stdout);#endif    int p;    while(~scanf("%d%d",&n,&p) && n+p)    {        nct = 0;        memset(T,0,sizeof(T));        memset(head,-1,sizeof(head));        for(int i = 0; i < n-1; i++)        {            int a,b;//            scan_d(a); scan_d(b);            scanf("%d%d",&a,&b);            add_edge(a,b);        }        dfs(p,-1);        for(int i = 1; i <= n; i++)            printf("%d%c",dp[i],(i==n)?'\n':' ');    }    return 0;}

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