POJ 3667
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 9671 | Accepted: 4145 |
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 Di (1 ≤ Di ≤ N) and approach the front desk to check in. Each group i requests a set of Di contiguous rooms from Canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbers r..r+Di-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 Xi and Di which specify the vacating of rooms Xi ..Xi +Di-1 (1 ≤ Xi ≤ N-Di+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 Di (b) Three space-separated integers representing a check-out: 2, Xi, and Di
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
10 6 1 3 1 3 1 3 1 3 2 5 5 1 6
Sample Output
1 4 7 0 5
Source
2 a b: 将[a,a+b-1]的房间清空
思路:记录区间中最长的空房间
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
using namespace std;
//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 50000 + 500;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
#define mod 1000000007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pi;
///#pragma comment(linker, "/STACK:102400000,102400000")
int lsum[MAXN <<2] , rsum[MAXN << 2], sum[MAXN << 2] , cover[MAXN << 2];
void pushup(int rt ,int len)
{
lsum[rt] = lsum[rt << 1];
rsum[rt] = rsum[rt << 1 | 1];
if(lsum[rt] == len - (len >> 1))lsum[rt] += lsum[rt << 1 | 1];
if(rsum[rt] == len >> 1)rsum[rt] += rsum[rt << 1];
sum[rt] = max(lsum[rt << 1 | 1] + rsum[rt << 1] , max(sum[rt << 1] , sum[rt << 1 | 1]));
}
void pushdown(int rt , int len)
{
if(cover[rt] != -1)
{
cover[rt << 1] = cover[rt << 1| 1] = cover[rt];
sum[rt << 1] = lsum[rt << 1] = rsum[rt << 1] = cover[rt] == 1? 0:len - (len >> 1);
sum[rt << 1 | 1] = lsum[rt << 1 | 1] = rsum[rt << 1 | 1] = cover[rt] == 1? 0:(len >> 1);
cover[rt] = -1;
}
}
void build(int l , int r, int rt)
{
sum[rt] = lsum[rt] = rsum[rt] = r - l + 1;
cover[rt] = -1;
if(l == r)return ;
int mid = (l + r) >> 1;
build(lson) , build(rson);
}
void update(int L ,int R , int c , int l , int r ,int rt)
{
if(L <= l&& r <= R)
{
sum[rt] = lsum[rt] = rsum[rt] = c ? 0 : r - l + 1;
cover[rt] = c;
return;
}
pushdown(rt , r - l + 1);
int mid = (l + r) >> 1;
if(L <= mid)update(L ,R , c , lson);
if(mid < R)update(L ,R , c , rson);
pushup(rt , r - l + 1);
}
int query(int w , int l , int r , int rt)
{
if(l == r)return l;
pushdown(rt , r - l + 1);
int mid = (l + r) >> 1;
if(sum[rt << 1] >= w)return query(w , lson);
else if(rsum[rt << 1] + lsum[rt << 1 | 1] >= w)return mid - rsum[rt << 1] + 1;
else return query(w , rson);
}
int main()
{
//ios::sync_with_stdio(false);
#ifdef Online_Judge
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif // Online_Judge
int n , m , a , b , op;
while(~scanf("%d%d" , &n , &m))
{
build(1 , n , 1);
while(m--)
{
scanf("%d" , &op);
if(op == 1)
{
scanf("%d", &a);
if(sum[1] < a)printf("0\n");
else
{
int p = query(a ,1 , n , 1);
printf("%d\n" , p);
update(p , p + a - 1 , 1 , 1 , n , 1);
}
}
else
{
scanf("%d%d" ,&a , &b);
update(a , b + a - 1 , 0 , 1 , n , 1);
}
}
}
return 0;
}