码迷,mamicode.com
首页 > 其他好文 > 详细

HDU 2871 Memory Control(线段树·区间合并·Vector)

时间:2015-08-16 23:15:37      阅读:97      评论:0      收藏:0      [点我收藏+]

标签:acm   线段树   

题意  模拟内存申请  有n个内存单元  有以下4种操作

  Reset  将n个内存单元全部清空

  New x  申请一个长度为x的连续内存块  申请成功就输出左端

  Free x  将x所在的内存块空间释放  释放成功输出释放的内存始末位置

  Get x  输出第x个内存块的起始位置

Reset 和 New 都是基本的区间合并知识  比较简单  Free和Get需要知道内层块的位置  所以我们在New的时候要将申请的内存块的始末位置都存起来  两个内层块不会相交  这样就能通过二分找到需要的内层块了  Free后要删去对应的内存块   用vector是为了使Get操作能在O(1)时间内完成  用List的话插入删除快一些但是Get就慢了  所以随便用什么来保存被占的内层块  相对来说vector比较容易实现

#include <cstdio>
#include <vector>
#include <algorithm>
#define lc p<<1, s, mid
#define lp p<<1
#define rc p<<1|1, mid + 1, e
#define rp p<<1|1
#define mid ((s+e)>>1)
using namespace std;
const int N = 50005;
int len[N << 2], lle[N << 2], lri[N << 2], setv[N << 2];

struct Block //内层块结构体
{
    int l, r;
    bool operator< (const Block &b) const{
        return l < b.l;
    }
} block;
vector<Block> vb;
vector<Block>::iterator it;

void pushup(int p, int s, int e)
{
    len[p] = max(len[lp], len[rp]);
    len[p] = max(len[p], lri[lp] + lle[rp]);
    lle[p] = lle[lp], lri[p] = lri[rp];
    if(lle[p] == mid - s + 1) lle[p] += lle[rp];
    if(lri[p] == e - mid)  lri[p] += lri[lp];
}

void pushdown(int p, int s, int e)
{
    if(setv[p] == -1) return;
    setv[lp] = setv[rp] = setv[p];
    len[lp] = lle[lp] = lri[lp] = setv[p] * (mid - s + 1);
    len[rp] = lle[rp] = lri[rp] = setv[p] * (e - mid);
    setv[p] = -1;
}

void build(int p, int s, int e)
{
    setv[p] = -1;
    if(s == e)
    {
        len[p] = lle[p] = lri[p] = 1;
        return;
    }
    build(lc);
    build(rc);
    pushup(p, s, e);
}

void update(int p, int s, int e, int l, int r, int v)
{
    if(l <= s && e <= r)
    {
        setv[p] = v;
        len[p] = lle[p] = lri[p] = v * (e - s + 1);
        return;
    }
    pushdown(p, s, e);
    if(l <= mid) update(lc, l, r, v);
    if(r > mid) update(rc, l, r, v);
    pushup(p, s, e);
}

int query(int p, int s, int e, int v)
{
    if(s == e) return s;
    pushdown(p, s, e);
    if(len[lp] >= v) return query(lc, v);
    if(lri[lp] + lle[rp] >= v) return mid + 1 - lri[lp];
    return query(rc, v);
}

int main()
{
    int n, m, x, p;
    char op[10];
    while(~scanf("%d%d", &n, &m))
    {
        build(1, 1, n);
        vb.clear();
        while(m--)
        {
            scanf("%s", op);

            if(op[0] == 'R') //Reset
            {
                update(1, 1, n, 1, n, 1); //用build会TLE
                vb.clear();
                puts("Reset Now");
            }

            if(op[0] == 'N')  //New
            {
                scanf("%d", &x);
                if(len[1] < x) puts("Reject New");
                else
                {
                    p = query(1, 1, n, x);
                    printf("New at %d\n", p);
                    update(1, 1, n, p, p + x - 1, 0);

                    block.l = p;
                    block.r = p + x - 1;
                    it = upper_bound(vb.begin(), vb.end(), block);
                    //找到第一个起点大于p的位置 确保插入后还是升序的
                    vb.insert(it, block);
                }
            }

            if(op[0] == 'G')  //Get
            {
                scanf("%d", &x);
                if(x > vb.size()) puts("Reject Get");
                else printf("Get at %d\n", vb[x - 1].l);
            }

            if(op[0] == 'F') //Free
            {
                scanf("%d", &x);
                block.l = block.r = x;
                it = upper_bound(vb.begin(), vb.end(), block);
                int i = it - vb.begin() - 1;
                if(i < 0 || vb[i].r < x) puts("Reject Free");
                else
                {
                    printf("Free from %d to %d\n", vb[i].l, vb[i].r);
                    update(1, 1, n, vb[i].l, vb[i].r, 1);
                    vb.erase(--it);
                }
            }
        }
        puts("");
    }
    return 0;
}

Memory Control



Problem Description
Memory units are numbered from 1 up to N.
A sequence of memory units is called a memory block. 
The memory control system we consider now has four kinds of operations:
1.  Reset Reset all memory units free.
2.  New x Allocate a memory block consisted of x continuous free memory units with the least start number
3.  Free x Release the memory block which includes unit x
4.  Get x Return the start number of the xth memory block(Note that we count the memory blocks allocated from left to right)
Where 1<=x<=N.You are request to find out the output for M operations. 
 

Input
Input contains multiple cases.
Each test case starts with two integer N,M(1<=N,M<=50000) ,indicating that there are N units of memory and M operations.
Follow by M lines,each line contains one operation as describe above.
 

Output
For each “Reset” operation, output “Reset Now”.
For each “New” operation, if it’s possible to allocate a memory block,
output “New at A”,where Ais the least start number,otherwise output “Reject New”.
For each “Free” operation, if it’s possible to find a memory block occupy unit x,
output “Free from A to B”,where A and B refer to the start and end number of the memory block,otherwise output “Reject Free”.
For each “Get” operation, if it’s possible to find the xth memory blocks,
output “Get at A”,where A is its start number,otherwise output “Reject Get”.
Output one blank line after each test case.
 

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

Sample Output
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
 

版权声明:本文为博主原创文章,未经博主允许不得转载。

HDU 2871 Memory Control(线段树·区间合并·Vector)

标签:acm   线段树   

原文地址:http://blog.csdn.net/acvay/article/details/47706629

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!