线段树

Online Judge

Problem Set

Authors

Online Contests

User

Web Board
Home Page
F.A.Qs
Statistical Charts

Current Contest
Past Contests
Scheduled Contests
Award Contest

Language:

Hotel

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 12496		Accepted: 5378

Description

The cows are journeying north to Thunder Bay in Canada to gain cultural enrichment and enjoy a vacation on the sunny shores of Lake Superior. Bessie, ever the competent travel agent, has named the Bullmoose Hotel on famed Cumberland Street as their vacation residence. This immense hotel has N (1 ≤ N ≤ 50,000) rooms all located on the same side of an extremely long hallway (all the better to see the lake, of course).

The cows and other visitors arrive in groups of size D_i (1 ≤ D_i ≤ N) and approach the front desk to check in. Each group i requests a set of D_i contiguous rooms from Canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbers r..r+D_i-1 if they are available or, if no contiguous set of rooms is available, politely suggests alternate lodging. Canmuu always chooses the value of r to be the smallest possible.

Visitors also depart the hotel from groups of contiguous rooms. Checkout i has the parameters X_i and D_i which specify the vacating of rooms X_i ..X_i +D_i-1 (1 ≤ X_i ≤ N-D_i+1). Some (or all) of those rooms might be empty before the checkout.

Your job is to assist Canmuu by processing M (1 ≤ M < 50,000) checkin/checkout requests. The hotel is initially unoccupied.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Line i+1 contains request expressed as one of two possible formats: (a) Two space separated integers representing a check-in request: 1 and D_i(b) Three space-separated integers representing a check-out: 2, X_i, and D_i

Output

* Lines 1.....: For each check-in request, output a single line with a single integer r, the first room in the contiguous sequence of rooms to be occupied. If the request cannot be satisfied, output 0.

Sample Input

Sample Output

Source

USACO 2008 February Gold

[Submit] [Go Back] [Status] [Discuss]

Home Page Go Back To top

//建树：按照二叉树

//结点信息:

lsum:保存的是以L为起点往右的最长连续空位

rsum:保存的是以R为起点往左的最长连续空位

sum:保存的是区间[L, R]中的最长连续空位

//操作信息：

0:本节点区间均为空房

1:本节点区间均已入住

-1:未进行任何操作

#include <cstdio>
#include <iostream>

using namespace std;

struct Node{
    int l, r, lsum, rsum, sum;
    int s;
}nodes[500000];

int max(int a, int b){
    return a>b?a:b;
}

int maxx(int a, int b, int c){
     int x1 = max(a, b);
     int x2 = max(b,c);
     return max(x1,x2);
}

void build(int o, int L, int R){
     nodes[o].l = L;
     nodes[o].r = R;
     int Len = R-L+1;
     nodes[o].lsum = nodes[o].rsum = nodes[o].sum = Len;
     nodes[o].s  = -1;

     if(L!=R){
        int M = (L+R)/2;
        build(2*o, L, M);
        build(2*o+1, M+1, R);
     }
}

void push_down(int o){
    int Lc = 2*o, Rc = 2*o+1;
    if(nodes[o].s >= 0){
        nodes[Lc].s = nodes[Rc].s = nodes[o].s;
        if(nodes[o].s){
            nodes[Lc].lsum = nodes[Lc].rsum = nodes[Lc].sum  = 0;
            nodes[Rc].lsum = nodes[Rc].rsum = nodes[Rc].sum  = 0;
        }
        else{
            nodes[Lc].lsum = nodes[Lc].rsum = nodes[Lc].sum  = nodes[Lc].r - nodes[Lc].l + 1;
            nodes[Rc].lsum = nodes[Rc].rsum = nodes[Rc].sum = nodes[Rc].r - nodes[Rc].l + 1;
        }
        nodes[o].s = -1;
    }
}

void push_up(int o){
     int Lc = 2*o, Rc = 2*o+1;
     int Len1 = nodes[Lc].r - nodes[Lc].l + 1;
     int Len2 = nodes[Rc].r - nodes[Rc].l + 1;

     if(nodes[Lc].lsum == Len1) nodes[o].lsum = Len1+nodes[Rc].lsum;
     else nodes[o].lsum = nodes[Lc].lsum;

     if(nodes[Rc].rsum == Len2) nodes[o].rsum = Len2 + nodes[Lc].rsum;
     else nodes[o].rsum = nodes[Rc].rsum;

     nodes[o].sum = maxx(nodes[Lc].sum, nodes[Rc].sum, (nodes[Lc].rsum + nodes[Rc].lsum));
}

void updata(int o, int y1, int y2, int cnt){
     int Lc = 2*o, Rc = 2*o+1;
     int L = nodes[o].l;
     int R = nodes[o].r;
     int Len = R-L+1;
     if(y1<=L && y2>=R){
        nodes[o].s = cnt;
        if(cnt)  nodes[o].lsum = nodes[o].rsum = nodes[o].sum = 0;
        else     nodes[o].lsum = nodes[o].rsum = nodes[o].sum = Len;
     }
     else{
        push_down(o);
        int M = (L + R)/2;
        if(y2<=M) updata(Lc,y1,y2,cnt);
        else if(y1>M) updata(Rc, y1, y2, cnt);
        else{
            updata(Lc, y1, M, cnt);
            updata(Rc, M+1, y2, cnt);
        }
        push_up(o);           //»ØËÝ¸üÐÂo
     }
}

int Search(int o, int len){
    int Lc = 2*o, Rc = 2*o+1;
    int L = nodes[o].l;
    int R = nodes[o].r;

    push_down(o);
    if(nodes[o].sum < len) return 0;
    if(L == R) return L;
    if(nodes[Lc].sum >= len){
        return Search(Lc, len);
    }
    else if((nodes[Lc].rsum + nodes[Rc].lsum) >= len){
        int lef = (L+R)/2 - nodes[Lc].rsum + 1;
        int rig = lef + len - 1;
        return lef;
    }
    else return Search(Rc, len);
}



int main(){
    int N,M;
    scanf("%d%d",&N, &M);
    build(1,1,N);
    while(M--){
        int cas;
        scanf("%d",&cas);
        if(cas == 1){
            int len;
            scanf("%d",&len);
            int ans = Search(1,len);
            printf("%d
",ans);
            if(ans) updata(1,ans,ans+len-1,1);
        }
        else{
            int b,len;
            scanf("%d%d",&b,&len);
            updata(1,b,b+len-1,0);
        }
    }
    return 0;
}