HDU 4614 (线段树)
2015-12-01 21:44
323 查看
Vases and Flowers
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 2615 Accepted Submission(s): 1018
Problem Description
Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to N-1. When she receive some flowers, she will try to put them in the vases, one flower in one vase. She randomly choose the vase A and try to put a flower in
the vase. If the there is no flower in the vase, she will put a flower in it, otherwise she skip this vase. And then she will try put in the vase A+1, A+2, ..., N-1, until there is no flower left or she has tried the vase N-1. The left flowers will be discarded.
Of course, sometimes she will clean the vases. Because there are too many vases, she randomly choose to clean the vases numbered from A to B(A <= B). The flowers in the cleaned vases will be discarded.
Input
The first line contains an integer T, indicating the number of test cases.
For each test case, the first line contains two integers N(1 < N < 50001) and M(1 < M < 50001). N is the number of vases, and M is the operations of Alice. Each of the next M lines contains three integers. The first integer of one line is K(1 or 2). If K
is 1, then two integers A and F follow. It means Alice receive F flowers and try to put a flower in the vase A first. If K is 2, then two integers A and B follow. It means the owner would like to clean the vases numbered from A to B(A <= B).
Output
For each operation of which K is 1, output the position of the vase in which Alice put the first flower and last one, separated by a blank. If she can not put any one, then output 'Can not put any one.'. For each operation of which K is 2, output the number
of discarded flowers.
Output one blank line after each test case.
Sample Input
2 10 5 1 3 5 2 4 5 1 1 8 2 3 6 1 8 8 10 6 1 2 5 2 3 4 1 0 8 2 2 5 1 4 4 1 2 3
Sample Output
[pre]3 7 2 1 9 4 Can not put any one. 2 6 2 0 9 4 4 5 2 3 [/pre]
裸的线段树,查找的时候可以直接通过节点上面的信息找到第一个和最后一个空瓶子.
#include <bits/stdc++.h> using namespace std; #define maxn 51111 #define pl c<<1 #define pr (c<<1)|1 #define lson tree[c].l,tree[c].mid,c<<1 #define rson tree[c].mid+1,tree[c].r,(c<<1)|1 #define Clear clear #define find Find struct node { int l, r, mid; int sum; bool full;//全满标记 bool em; //全空标记 }tree[maxn<<4]; int n, m; void build_tree (int l, int r, int c) { tree[c].l = l, tree[c].r = r, tree[c].mid = (l+r)>>1, tree[c].sum = 0; tree[c].em = 1, tree[c].full = 0; if (l == r) return ; build_tree (lson); build_tree (rson); return ; } void push_down (int c) { if (tree[c].l == tree[c].r) return ; if (tree[c].em) { tree[pl].sum = tree[pl].full = 0; tree[pl].em = 1; tree[pr].sum = tree[pr].full = 0; tree[pr].em = 1; } else if (tree[c].full) { tree[pl].sum = (tree[pl].r-tree[pl].l+1); tree[pr].sum = (tree[pr].r-tree[pr].l+1); tree[pl].full = tree[pr].full = 1; tree[pl].em = tree[pr].em = 0; } tree[c].em = tree[c].full = 0; return ; } void push_up (int c) { if (tree[c].l == tree[c].r) return ; tree[c].sum = tree[pl].sum + tree[pr].sum; tree[c].full = tree[pl].full&tree[pr].full; tree[c].em = tree[pl].em&tree[pr].em; return ; } int sum (int l, int r, int c, int x, int y) { if (y < x) return 0; push_down (c); int ans = 0; if (l == x && r == y) { return tree[c].sum; } else if (tree[c].mid >= y) { return sum (lson, x, y); } else if (tree[c].mid < x) { return sum (rson, x, y); } else { return sum (lson, x, tree[c].mid) + sum (rson, tree[c].mid+1, y); } } void clear (int l, int r, int c, int x, int y) { //清空花瓶 if (y < x) return ; push_down (c); if (x == l && y == r) { tree[c].em = 1; tree[c].full = 0; tree[c].sum = 0; return ; } else if (tree[c].mid >= y) { clear (lson, x, y); } else if (tree[c].mid < x) { clear (rson, x, y); } else { clear (lson, x, tree[c].mid); clear (rson, tree[c].mid+1, y); } push_up (c); } int find (int l, int r, int c, int x) { push_down (c); if (l == r) return l; int all = tree[pl].r-tree[pl].l+1-tree[pl].sum; //左儿子的空瓶子 if (all >= x) { return find (lson, x); } else return find (rson, x-all); } void add (int l, int r, int c, int x, int y) { push_down (c); if (l == x && r == y) { tree[c].sum = r-l+1; tree[c].full = 1; tree[c].em = 0; return ; } else if (tree[c].mid >= y) { add (lson, x, y); } else if (tree[c].mid < x) { add (rson, x, y); } else { add (lson, x, tree[c].mid); add (rson, tree[c].mid+1, y); } push_up (c); } void solve (int x, int y) { int ans = sum (0, n-1, 1, x, n-1); if (n-x-ans == 0) { printf ("Can not put any one.\n"); return ; } int l = 0, r = 0; int pre = n-x-ans; ans = sum (1, n, 1, 0, x-1); ans = x-ans; //x之前有ans个空花瓶 l = find (0, n-1, 1, ans+1); //找到第ans+1个空花瓶的位置 if (!r) r = find (0, n-1, 1, min (ans+y, ans+pre)); //找到第ans+y个空花瓶的位置 printf ("%d %d\n", l, r); add (0, n-1, 1, l, r); } void debug (int c) { cout << "l:" << tree[c].l << " r:" << tree[c].r << " sum:" << tree[c].sum << endl; if (tree[c].l == tree[c].r) return ; debug (pl); debug (pr); return ; } int main () { //freopen ("in", "r", stdin); int t; scanf ("%d", &t); while (t--) { scanf ("%d%d", &n, &m); build_tree (0, n-1, 1); for (int i = 1; i <= m; i++) { int op; scanf ("%d", &op); if (op == 1) { int x, y; scanf ("%d%d", &x, &y); solve (x, y); } else if (op == 2) { int x, y; scanf ("%d%d", &x, &y); printf ("%d\n", sum (0, n-1, 1, x, y)); clear (0, n-1, 1, x, y); } } printf ("\n"); } return 0; }
相关文章推荐
- 初识RMAN 1:数据文件和控制文件位置参数配置
- ruby的hash学习笔记例: 将字符串文本中的单词存放在map中
- 求n阶勒让德多项式
- 深入分析JavaWeb Item3 -- Tomcat服务器学习和使用2
- RunTime
- 信息安全系统设计基础实验四
- 转: servlet中的service, doGet, doPost方法的区别和联系
- RHEL6忘记root密码的解决办法
- 比较好的博客论文保存
- 1020. Tree Traversals (25)——PAT (Advanced Level) Practise
- 用递归方法求 f(n)
- lodash用法系列(3),使用函数
- hdoj Kingdom of Black and White 5583 (模拟&&DP) 好题
- ios 开发中跟绘图相关的CGFloat,CGPoint,CGSize,CGRect
- LeetCode 144 Binary Tree Preorder Traversal
- openssl运行出现 no OPENSSL_Applink解决办法
- ps aux|grep memcached Linux下查看memcachedzhe个进程
- poj1637 Sightseeing tour 混合图欧拉回路判定
- windows下配置nginx+php环境
- 有序回文数