Vases and Flowers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1336    Accepted Submission(s): 532

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 Input2 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 3Sample Output[pre]3 7 2 1 9 4 Can not put any one. 2 6 2 0 9 4 4 5 2 3 [/pre]Source

題意：n個花瓶從0到n-1，剛開始都是空的，有兩種操作，第一種操作是從A位置開始放放F只花，每個花瓶只能放一束花，如果花瓶內已經有花則跳過，直到F只花放完或者放到n-1只花瓶時結束，輸出開始放花的位置和結束放花的位置，如果從A開始沒有空花瓶，輸出一句話，第二種操作是，輸出區間【A,B】的有花的花瓶數，並把有花的花瓶清空。

思路：線段樹題，第二種操作很好處理，區間查詢之後再區間更新，把區間置為0，使用了lazy，第一種操作比較麻煩，如果我們直到了開始放花的位置和結束放花的位置，直接區間更新，全置為1就行了，如何得到這兩個位置呢，方法是二分，兩次二分，首先二分起始位置，然後在二分結束位置。

#include <stdio.h>
#include <algorithm>

using namespace std;

#define lson num << 1
#define rson num << 1 | 1
#define maxn 50010
#define mod 10007

typedef long long ll;

struct node
{
    int l,r;
    int sum;
    int lazy;
}segTree[maxn << 2];
int ans;
void pushup(int num)
{
    segTree[num].sum = segTree[lson].sum + segTree[rson].sum;
}
void build(int num,int l,int r)
{
    segTree[num].l = l;
    segTree[num].r = r;
    segTree[num].lazy = -1;
    if(l == r){
        segTree[num].sum = 0;
        return;
    }
    int mid = (segTree[num].l + segTree[num].r) >> 1;
    build(lson,l,mid);
    build(rson,mid + 1,r);
    pushup(num);
}
void pushdown(int num)
{
    if(segTree[num].lazy != -1){
        segTree[lson].lazy = segTree[rson].lazy = segTree[num].lazy;
        segTree[lson].sum = (segTree[lson].r - segTree[lson].l + 1) * segTree[num].lazy;
        segTree[rson].sum = (segTree[rson].r - segTree[rson].l + 1) * segTree[num].lazy;
        segTree[num].lazy = -1;
    }
}
void update(int num,int l,int r,int v)
{
    if(segTree[num].l == l && segTree[num].r == r){
        segTree[num].sum = (r - l + 1) * v;
        segTree[num].lazy = v;
        return;
    }
    pushdown(num);
    int mid = (segTree[num].l + segTree[num].r) >> 1;
    if(r <= mid){
        update(lson,l,r,v);
    }
    else if(l > mid){
        update(rson,l,r,v);
    }
    else{
        update(lson,l,mid,v);
        update(rson,mid + 1,r,v);
    }
    pushup(num);
}
int query(int num,int l,int r)
{
    if(segTree[num].l == l && segTree[num].r == r){
        return segTree[num].sum;
    }
    pushdown(num);
    int mid = (segTree[num].l + segTree[num].r) >> 1;
    int ans = 0;
    if(r <= mid){
        ans += query(lson,l,r);
    }
    else if(l > mid){
        ans += query(rson,l,r);
    }
    else{
        ans += query(lson,l,mid);
        ans += query(rson,mid + 1,r);
    }
    pushup(num);
    return ans;
}
int main(void)
{
    int T;
    scanf("%d",&T);
    while(T--){
        int n,m;
        scanf("%d%d",&n,&m);
        build(1,1,n);
        while(m--){
            int op;
            int l,r;
            int p,num;
            scanf("%d",&op);
            if(op == 1){
                scanf("%d%d",&p,&num);
                if(query(1,p + 1,n) == (n - (p + 1) + 1)){
                    printf("Can not put any one.\n");
                }
                else{
                    int L = p + 1,R = n;
                    int mid;
                    int pos1 = n;
                    while(L <= R){
                        int mid = (L + R) >> 1;
                        if(mid - (p + 1) + 1 > query(1,p + 1,mid)){
                            pos1 = min(pos1,mid);
                            R = mid - 1;
                        }
                        else{
                            L = mid + 1;
                        }
                    }
                    int temp = n - pos1 + 1 - query(1,pos1,n);
                    int pos2 = n;
                    if(num >= temp){
                        num = temp;
                    }
                    L = pos1,R = n;
                    while(L <= R){
                        int mid = (L + R) >> 1;
                        if(mid - pos1 + 1 - query(1,pos1,mid) >= num){
                            pos2 = min(pos2,mid);
                            R = mid - 1;
                        }
                        else{
                            L = mid + 1;
                        }
                    }
                    printf("%d %d\n",pos1 - 1,pos2 - 1);
                    update(1,pos1,pos2,1);
                }

            }
            else{
                scanf("%d%d",&l,&r);
                printf("%d\n",query(1,l + 1,r + 1));
                update(1,l + 1,r + 1,0);
            }
        }
        printf("\n");
    }
    return 0;
}