HDU 4614 (线段树)

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 
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;
}