hdoj 2871 Memory Control 【线段树lazy + 区间合并 + 二维vector的查询、插入、删除】

6 10
New 2
New 5
New 2
New 2
Free 3
Get 1
Get 2
Get 3
Free 3
Reset

New at 1
Reject New
New at 3
New at 5 
Free from 3 to 4
Get at 1
Get at 5
Reject Get
Reject Free
Reset Now

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define MAXN 50000+10
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
struct Tree
{
    int l, r, len;
    int lsum, rsum;
    int sum;
    int lazy;
};
Tree tree[MAXN<<2];
void PushUp(int o)
{
    tree[o].lsum = tree[ll].lsum;
    tree[o].rsum = tree[rr].rsum;
    if(tree[ll].lsum == tree[ll].len)//左半区间 可以全部容纳
        tree[o].lsum += tree[rr].lsum;
    if(tree[rr].rsum == tree[rr].len)
        tree[o].rsum += tree[ll].rsum;
    //更新最值 (左子树最值, 右子树最值, 左子树右连续长度 + 右子树左连续长度)
    tree[o].sum = max(max(tree[ll].sum, tree[rr].sum), tree[ll].rsum + tree[rr].lsum);
}
void build(int o, int l, int r)
{
    tree[o].l = l; tree[o].r = r;
    tree[o].len = tree[o].lsum = tree[o].rsum = tree[o].sum = r - l + 1;
    tree[o].lazy = -1;
    if(l == r)
        return ;
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
}
void PushDown(int o)
{
    if(tree[o].lazy != -1)
    {
        tree[ll].lazy = tree[rr].lazy = tree[o].lazy;
        tree[ll].lsum = tree[ll].rsum = tree[ll].sum = tree[o].lazy ? 0 : tree[ll].len;
        tree[rr].lsum = tree[rr].rsum = tree[rr].sum = tree[o].lazy ? 0 : tree[rr].len;
        tree[o].lazy = -1;
    }
}
int query(int need, int o)
{
    if(tree[o].l == tree[o].r)
        return tree[o].l;
    PushDown(o);//下传 lazy标记
    int mid = (tree[o].l + tree[o].r) >> 1;
    //查询顺序 不能变
    if(need <= tree[ll].sum)
        return query(need, ll);
    else if(need <= tree[ll].rsum + tree[rr].lsum)
        return mid - tree[ll].rsum + 1;
    else if(need <= tree[rr].sum)
        return query(need, rr);
}
void update(int L, int R, int o, int v)
{
    if(L <= tree[o].l && R >= tree[o].r)
    {
        tree[o].lsum = tree[o].rsum = tree[o].sum = v ? 0 : tree[o].len;
        tree[o].lazy = v;
        return ;
    }
    PushDown(o);
    int mid = (tree[o].l + tree[o].r) >> 1;
    if(R <= mid)
        update(L, R, ll, v);
    else if(L > mid)
        update(L, R, rr, v);
    else
    {
        update(L, mid, ll, v);
        update(mid+1, R, rr, v);
    }
    PushUp(o);//维护 上区间
}
struct Node
{
    int s, e;//block 起点 终点
};
vector<Node> G;
vector<Node> ::iterator it;
bool cmp(Node a, Node b)
{
    return a.s < b.s;//起点升序排列
}
int main()
{
    int n, m;
    while(scanf("%d%d", &n, &m) != EOF)
    {
        G.clear();//初始化
        build(1, 1, n);//建树
        while(m--)
        {
            char op[10];
            int x, pos;
            scanf("%s", op);
            if(op[0] == 'N')
            {
                scanf("%d", &x);
                if(tree[1].sum < x)//不能容纳内存
                    printf("Reject New\n");
                else
                {
                    pos = query(x, 1);//查询最小起点
                    printf("New at %d\n", pos);
                    update(pos, pos+x-1, 1, 1);//占据 连续的空间
                    Node E = {pos, pos+x-1};//容器存储
                    it = upper_bound(G.begin(), G.end(), E, cmp);//查询起点大于E.s的第一个 block
                    G.insert(it, E);//插入
                }
            }
            else if(op[0] == 'F')
            {
                scanf("%d", &x);//去掉x所在的 block
                Node E = {x, x};
                it = upper_bound(G.begin(), G.end(), E, cmp);//查询起点大于E.s的第一个 block
                pos = it - G.begin() - 1;//得到需要的block在vector里面的下标
                //判断是否在 该block里面
                if(pos < 0 || G[pos].s > x || G[pos].e < x)
                    printf("Reject Free\n");
                else
                {
                    update(G[pos].s, G[pos].e, 1, 0);
                    printf("Free from %d to %d\n", G[pos].s, G[pos].e);
                    G.erase(G.begin() + pos);
                }
            }
            else if(op[0] == 'G')
            {
                scanf("%d", &x);//第x个block
                x -= 1;
                if(x < 0 || x >= G.size())
                    printf("Reject Get\n");
                else
                    printf("Get at %d\n", G[x].s);
            }
            else
            {
                printf("Reset Now\n");
                G.clear();
                update(1, n, 1, 0);//释放所有内存
            }
        }
        printf("\n");
    }
    return 0;
}