HDU 4125 NlogN查找二叉树的生成

来源：互联网发布：安庆网络广播电视电话编辑：程序博客网时间：2024/06/05 02:07

题意：题目给定一颗二叉树，前序遍历二叉树的时候产生一个01的欧拉序列，输出某个01串在此序列中出现了几次。

总结：真是弱。。。。我居然模拟查找二叉树的生成，这可能退化到N^2。。。，引用了一个结论：

http://www.cppblog.com/hanfei19910905/archive/2012/07/02/181144.html

nlogn的算法生成静态二叉树，mark一下：

#include <cstdio>#include <cstring>#include <algorithm>#include <set>#pragma comment(linker, "/STACK:36777216")using namespace std;const int N = 600050;char s[N << 1],p[N];int next[N],n,a[N],idx,lenp,lens,tot;int left[N],right[N],stack[N],top;bool vis[N];set<int> myset;void init(int key){    myset.clear();    top = 0;    for(int i = 1;i <= n;i ++)        left[i] = right[i] = -1;    myset.insert(key);}void insert(int key){    // printf("key:%d  ",key);    myset.insert(key);    set<int> ::iterator less,more;    more = less = myset.lower_bound(key);    if(less != myset.begin()){        less --;        if(right[*less] == -1) {right[*less] = key;            // printf("less:%d\n",*less);            return ;        }    }    more ++;    if(more != myset.end() && left[*more] == -1) left[*more] = key;    // printf("more:%d\n",*more);}void setNext(){    int j = 0 , k = -1;    next[j] = k;    while(j < lenp){        if(k == -1 || p[j] == p[k]){            if(p[j + 1] == p[k + 1]) next[++ j] = next[++ k];            else next[++ j] = ++k;        }        else k = next[k];    }}void dfs(int cur){    memset(vis,false,sizeof(vis));    top = 0;    stack[++ top] = cur;    while(!vis[cur]){        int u = stack[top];        s[lens ++] = '0' + (u & 1);        int v = left[u];        if(v != -1 && !vis[v]) {stack[++ top] = v;continue;}        v = right[u];        if(v != -1 && !vis[v]) {stack[++ top] = v; continue;}        top --;        vis[u] = 1;    }}int kmp(){    int i = 0,j = 0;    int Count = 0;    while(i < lens){        if(j == -1 || s[i] == p[j]){            i ++,j ++;            if(j == lenp) Count ++;        }        else j = next[j];    }    return Count;}int main(){    // freopen("input.txt","r",stdin);    int cas;    scanf("%d",&cas);    for(int t = 1;t <= cas;t ++){        scanf("%d",&n);        for(int i = 0;i < n;i ++)            scanf("%d",&a[i]);        scanf("%s",p);        init(a[0]);        for(int i = 1;i < n;i ++)            insert(a[i]);        // printf("LLLL:%d\n",left[5]);        lens = 0;        dfs(a[0]);        s[lens] = '\0';        lenp = strlen(p);        // printf("s:%s\n",s);        setNext();        int ans = kmp();        printf("Case #%d: %d\n",t,ans);    }    return 0;}