POJ 3667 HOTEL

/*
第一类信息
ll, rr, len
递推求解
第二类信息
state.表示整个子树的性质
*/
#include <stdio.h>
#include <algorithm>
#define L(x) ((x) << 1)
#define R(x) ((x) << 1 | 1)
using namespace std;

const int MAXN = 50002;
int N, M;
struct SegTree
{
    int l, r;
    int ll, rr;    //最左边，右边可用连续区间长度
    int len;    //最长连续可用房间长度
    int state;    //0 空 1 满 -1 不空不满
    int getMid(){
        return (l + r) >> 1;
    }
    int getLen(){
        return r - l + 1;
    }
    void init(int state1){
        if(!state1){    //房间空闲
            ll = rr = len = getLen();
            state = state1;
        } else {    //state = 1 区间房间全满
            ll = rr = len = 0;
            state = state1;
        }
    }
};
SegTree tree[MAXN << 2];

void bulid(int left, int right, int t){
    tree[t].l = left;
    tree[t].r = right;
    tree[t].len = tree[t].getLen();
    tree[t].ll = tree[t].rr = tree[t].len;
    tree[t].state = 0;
    if(left == right)
        return;
    int mid = tree[t].getMid();
    bulid(left, mid, L(t));
    bulid(mid + 1, right, R(t));
}

void update(int left, int right, int state, int t){
    if(tree[t].state == state){            //整个区间颜色和state相同 直接返回
        return;
    }
    if(left <= tree[t].l && right >= tree[t].r){        //包含此区间
        tree[t].init(state);                //更新第一类信息 
        return;
    }
    if(tree[t].state != -1){                //向下传递第二类信息
        tree[L(t)].state = tree[R(t)].state = tree[t].state;
        tree[L(t)].init(tree[L(t)].state);
        tree[R(t)].init(tree[R(t)].state);
    }
    int mid = tree[t].getMid();
    if(right <= mid){
        update(left, right, state, L(t));
    } else if(left > mid){
        update(left, right, state, R(t));
    } else {
        update(left, mid, state, L(t));
        update(mid + 1, right, state, R(t));
    }
    if(tree[L(t)].state == 0 && tree[R(t)].state == 0){
        tree[t].state = 0;
    } else if(tree[L(t)].state == 1 && tree[R(t)].state == 1 ){
        tree[t].state = 1;
    } else {
        tree[t].state = -1;
    }
    //第一类信息的维护
    tree[t].ll = tree[L(t)].ll + (tree[L(t)].state == 0 ? tree[R(t)].ll : 0);
    tree[t].rr = tree[R(t)].rr + (tree[R(t)].state == 0 ? tree[L(t)].rr : 0);
    tree[t].len = max(tree[L(t)].len , max(tree[R(t)].len , tree[L(t)].rr + tree[R(t)].ll));
}

int checkIn(int a, int t){
    if(tree[t].state == 0 && tree[t].len >= a){//如果当前区间长度满足 直接返回区间左端点
        return tree[t].l;
    }
    if(tree[t].state == 1){//如果全部是满的，那就不要向下查询了，直接返回0
        return 0;
    }
    if(tree[t].state == -1){            //分三种情况考虑 左边 中间 右边
        if(tree[L(t)].len >= a){
            return checkIn(a, L(t));
        } else if(tree[L(t)].rr + tree[R(t)].ll >= a){
            return tree[t].getMid() - tree[L(t)].rr + 1;
        } else {
            return checkIn(a, R(t));
        }
    }
    return 0;
}

int main(){
    scanf("%d%d",&N, &M);
    int cmd, a, b;
    bulid(1, N, 1);
    for(int i = 0; i < M; ++i){
        scanf("%d",&cmd);
        if(cmd == 1){
            scanf("%d",&a);
            int ans = checkIn(a, 1);
            printf("%d\n",ans);
            if(ans){
                update(ans, ans + a - 1, 1, 1);    //1进 
            }
        } else {
            scanf("%d %d",&a, &b);
            update(a, a + b - 1, 0, 1);        //0 出 
        }
    }
    return 0;
}