• poj-3321-dfs序-线段树-邻接表


    思路:用邻接表存图,卡vector【这里被卡哭了QAQ】,用dfs遍历的顺序重新给节点编号,遍历时记录儿子数目。用dfs序建立线段树,change的时候单点更新,查询某子树上的苹果树即是查询该节点[i, i+childnum]这个区间的苹果数目,i指dfs序。

     总结:邻接表出边入边傻傻搞不清楚QAQ

    AC代码:

      1 #include <iostream>
      2 #include <cstdio>
      3 #include <vector>
      4 #include <cstring>
      5 #include <algorithm>
      6 using namespace std;
      7 #define maxn 200010
      8 #define lson l, m, rt<<1
      9 #define rson m+1, r, rt<<1|1
     10 struct node
     11 {
     12     int cnt, val, num;
     13 };
     14 int n, m;
     15 node arr[maxn];
     16 int u[maxn], v[maxn], next[maxn], first[maxn];
     17 bool vis[maxn];
     18 int sgt[maxn<<2];
     19 void init()
     20 {
     21     memset(vis, 0, sizeof(vis));
     22     for(int i = 0; i < n+10; i++)  first[i] = -1;
     23     int e = 0, i;
     24     for(i = 1; i < n; i++) {
     25         scanf("%d%d", &u[i], &v[i]);
     26         next[i] = first[u[i]];
     27         first[u[i]] = i;
     28         u[n-1+i] = v[i];
     29         v[n-1+i] = u[i];
     30         next[n-1+i] = first[u[n-1+i]];
     31         first[u[n-1+i]] = n-1+i;
     32     }
     33 }
     34 int dfs(int i, int &num)
     35 {//cout<<i<<" - "<<num<<endl;
     36     arr[i].val = 1;
     37     arr[i].cnt = 0;
     38     arr[i].num = num;
     39     vis[i] = 1;
     40     for(int j = first[i]; j != -1; j = next[j]) {
     41         if(!vis[v[j]]) { num++; arr[i].cnt += dfs(v[j], num);  }
     42     }
     43     return arr[i].cnt+1;
     44 }
     45 void push_up(int rt)
     46 {
     47     sgt[rt] = sgt[rt<<1] + sgt[rt<<1|1];
     48 }
     49 void build(int l, int r, int rt)
     50 {
     51     sgt[rt] = 1;
     52     if(l == r) return;
     53     int m = (r+l)>>1;
     54     build(lson);
     55     build(rson);
     56     push_up(rt);
     57 }
     58 void change(int l, int r, int rt, int pos)
     59 {
     60     if(l == r) {
     61         if(sgt[rt] == 1) sgt[rt] = 0;
     62         else sgt[rt] = 1;
     63         return;
     64     }
     65     int m = (r+l)>>1;
     66     if(pos <= m) change(lson, pos);
     67     else change(rson, pos);
     68     push_up(rt);
     69 }
     70 int query(int l, int r, int rt, int L, int R )
     71 {
     72     if(L <= l && r <= R) {
     73         return sgt[rt];
     74     }
     75     int res = 0;
     76     int m = (l+r)>>1;
     77     if(L <= m) res += query(lson, L, R);
     78     if(m < R) res += query(rson, L, R);
     79     return res;
     80 }
     81 void work()
     82 {
     83     if(n == 1) {
     84         int x = 1;
     85         scanf("%d", &m);
     86         char re; int k;
     87         while (m--) {
     88             getchar();
     89             scanf("%c%d", &re, &k);
     90             if(re == 'Q') {
     91                 printf("%d
    ", x);
     92             }
     93             else {
     94                 if(x == 0) x = 1; else x = 0;
     95             }
     96         }
     97     }else {
     98         init();
     99         int xxx = 1;
    100         dfs(1,xxx);
    101         build(1, n, 1);
    102         char re; int k;
    103         scanf("%d", &m);
    104         for (int i = 0; i < m; i++) {
    105             getchar();
    106             scanf("%c%d", &re, &k);
    107             if(re == 'C') {
    108                 change(1, n, 1, arr[k].num);
    109             }
    110             else {
    111                 int res = query(1, n, 1, arr[k].num, arr[k].num+arr[k].cnt);
    112                 printf("%d
    ", res);
    113             }
    114         }
    115     }
    116 }
    117 int main()
    118 {
    119     while(scanf("%d", &n) != EOF && n) work();
    120     return 0;
    121 }
    122 /*
    123 7
    124 1 3
    125 1 4
    126 3 5
    127 3 6
    128 4 2
    129 4 7
    130 11
    131 Q 2
    132 Q 2
    133 Q 1
    134 C 3
    135 Q 3
    136 C 3
    137 Q 3
    138 C 4
    139 Q 7
    140 Q 4
    141 Q 2
    142 
    143 */
    View Code
  • 相关阅读:
    redis发布订阅
    redis学习笔记(面试题)
    redis安全 (error) NOAUTH Authentication required
    HDU3001 Travelling —— 状压DP(三进制)
    POJ3616 Milking Time —— DP
    POJ3186 Treats for the Cows —— DP
    HDU1074 Doing Homework —— 状压DP
    POJ1661 Help Jimmy —— DP
    HDU1260 Tickets —— DP
    HDU1176 免费馅饼 —— DP
  • 原文地址:https://www.cnblogs.com/ZiningTang/p/3962849.html
Copyright © 2020-2023  润新知