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hdu 2871 Memory Control——线�D�|��

Tue, 10 May 2011 14:13:00 GMT

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

#include <stdio.h>
#include <stdlib.h>
#include <vector>
#include <iostream>
#define maxn 51000
using namespace std;
inline int L(int th)
{
    return (th << 1);
}
inline int R(int th)
{
    return ((th << 1) | 1);
}
struct Tree
{
    int l, r;
    int cover, rely;
    int lm, rm, maxl;
    int getLen()
    {
        return r - l;
    }
    int mid()
    {
        return ((l + r) >> 1);
    }
}stree[maxn * 4];
void build(int l, int r, int th)
{
    //printf("%d %d\n", l, r);
    stree[th].l = l, stree[th].r = r;
    stree[th].cover = stree[th].rely = 0;
    stree[th].lm = stree[th].rm = stree[th].maxl = stree[th].getLen();
    if (l + 1 == r)
    {
        return;
    }
    int m = stree[th].mid();
    if (r <= m)
    {
        build(l, r, L(th));
    }
    else if (l >= m)
    {
        build(l, r, R(th));
    }
    else
    {
        build(l, m, L(th)), build(m, r, R(th));
    }
}
void update(int th, int kind)//rely操作更新
{
    stree[th].rely = 1;
    stree[th].cover = kind;
    if (stree[th].cover == 0)
    {
        stree[th].lm = stree[th].rm = stree[th].maxl = stree[th].getLen();
    }
    else
    {
        stree[th].lm = stree[th].rm = stree[th].maxl = 0;
    }
}
void insert(int l, int r, int th, int kind)
{
    //printf("insert %d, %d\n", l, r);
    if (stree[th].l == l && stree[th].r == r)
    {
        update(th, kind);
        return;
    }
    if (stree[th].rely)
    {
        stree[th].rely = 0;
        update(L(th), stree[th].cover);
        update(R(th), stree[th].cover);
    }
    stree[th].cover = -1;//讄��无效区域�Q�在�q�里体现不上用处�Q�不更新也行,��Z��习惯跟逻辑需�?nbsp;
    int m = stree[th].mid();
    if (r <= m)
    {
        insert(l, r, L(th), kind);
    }
    else if (l >= m)
    {
        insert(l, r, R(th), kind);
    }
    else
    {
        insert(l, m, L(th), kind);
        insert(m, r, R(th), kind);
    }
    stree[th].lm = stree[L(th)].lm, stree[th].rm = stree[R(th)].rm;
    if (stree[L(th)].lm == stree[L(th)].getLen())
    {
        stree[th].lm += stree[R(th)].lm;
    }
    if (stree[R(th)].rm == stree[R(th)].getLen())
    {
        stree[th].rm += stree[L(th)].rm;
    }
    stree[th].maxl = max(stree[L(th)].maxl, stree[R(th)].maxl);
    stree[th].maxl = max(stree[th].maxl, stree[L(th)].rm + stree[R(th)].lm);
}
int query(int th, int len)
{
    if (stree[th].l + 1 == stree[th].r)
    {
        if (stree[th].maxl >= len)
        {
            return stree[th].l;
        }
        return 0;
    }
    if (stree[th].rely)
    {
        stree[th].rely = 0;
        update(L(th), stree[th].cover);
        update(R(th), stree[th].cover);
    }
    if (stree[th].maxl < len)//该线�D�|��大空白段maxl��于len
    {
        return 0;
    }
    if (stree[L(th)].maxl >= len)//左孩子最大空白段满��len
    {
        return query(L(th), len);
    }
    else if(stree[L(th)].rm + stree[R(th)].lm >= len)//�׃��孩子的中间合�q�满��?nbsp;
    {
        return stree[L(th)].r - stree[L(th)].rm;
    }
    else//叛_��子满��?nbsp;
    {
        return query(R(th), len);
    }
}
struct Seg
{
    int left, right;
};
vector<Seg>list;
int bsc(int key)
{
    //查找key所在的位置的后一位，
    //如果不存在包含key的线�D�|��Q?br>    //1.key比第一个线�D�小�Q�返回的�?,0-1��q��?1
    //2.key比�v�Ҏ��大线�D�还大，则返回这个线�D늚�后一位，key可能被这个线�D�包含，可能不被�q�个�U�段包含�Q�取决于�q�个�U�段的右端点
    int l = 0, r = list.size() - 1, m;
    while (l <= r)
    {
        m = (l + r) >> 1;
        if (list[m].left > key)
        {
            r = m - 1;
        }
        else if (list[m].left <= key)//保证l在满��线�D늚�后一�?nbsp;
        {
            l = m + 1;
        }
    }
    return l;
}
void show()
{
    int i;
    for (i = 0; i < list.size(); i++)
    {
        printf("%d:%d %d\n", i, list[i].left, list[i].right);
    }printf("----\n");
}
int main()
{
    int len, x, k, n, m, id;
    Seg tmp;
    while (scanf("%d%d", &n, &m) - EOF)
    {
        build(1, n + 1, 1);
        char op[10];
        list.clear();
        while (m--)
        {
            scanf("%s", op);
            if (*op == 'N')
            {
                scanf("%d", &len);
                k = query(1, len);
                if (k)
                {
                    printf("New at %d\n", k);
                    insert(k, k + len, 1, 1);
                    tmp.left = k, tmp.right = k + len;
                    if (list.size() == 0)
                    {
                        list.push_back(tmp);
                    }
                    else
                    {
                        id = bsc(k);
                        list.insert(list.begin() + id, tmp);
                    }
                    //show();
                }
                else
                {
                    printf("Reject New\n");
                }
            }
            else if (*op == 'F')
            {
                scanf("%d", &x);
                k = bsc(x) - 1;
                if (k == -1 || list[k].right <= x)
                {
                    printf("Reject Free\n");
                }
                else
                {
                    printf("Free from %d to %d\n", list[k].left, list[k].right - 1);
                    insert(list[k].left, list[k].right, 1, 0);
                    list.erase(list.begin() + k, list.begin() + k + 1);
                }
            }
            else if (*op == 'G')
            {
                scanf("%d", &x);
                if (x <= list.size())
                {
                    printf("Get at %d\n", list[x-1].left);
                }
                else
                {
                    printf("Reject Get\n");
                }
            }
            else if (*op == 'R')
            {
                insert(1, n + 1, 1, 0);
                list.clear();
                printf("Reset Now\n");
            }
        }
        printf("\n");
    }
    return 0;
}
/*
3 100
N 1
N 1
N 1
F 1
N 1
F 2
N 1
F 3
N 2
*/

火碳�?/a> 2011-05-10 22:13 发表评论

Thu, 05 May 2011 04:14:00 GMT

敌兵布阵

Problem Description

C国的��d��头A国这�D�|��间正在进行军事演习，所以C国间谍头子Derek和他手下Tidy又开始忙乎了。A国在��岸�U�沿直线布置了N个工兵营�? Derek和Tidy的�Q务就是要监视�q�些工兵营地的活动情��c��由于采取了某种先进的监��手�D�，所以每个工兵营地的人数C国都掌握的一清二�?每个工兵�? 地的人数都有可能发生变动�Q�可能增加或减少若干人手,但这些都逃不�q�C国的监视�?br>�? 央情报局要研�I�敌人究竟演习什么战�?所以Tidy要随时向Derek汇报某一�D�连�l�的工兵营地一共有多少�?例如Derek�?“Tidy,马上汇报�W? 3个营地到�W?0个营地共有多��h!”Tidy��p��马上开始计��这一�D늚��M�h数�ƈ汇报。但敌兵营地的�h数经常变动，而Derek每次询问的段都不一��P��所以Tidy不得不每�ơ都一个一个营地的��L��Q�很快就�_��力尽了，Derek对Tidy的计��速度��来��不�?"你个死肥仔，��得�q�么慢，我炒你鱿�?” Tidy惻I��“你自己来��算看，�q�可真是一��篏人的工作!我恨不得你炒我鱿鱼呢!”无奈之下�Q�Tidy只好打电话向计算��Z��家Windbreaker�? �?Windbreaker��_��“死肥仔，叫你�q�x��做多点acm题和看多点算法书�Q�现在尝到苦果了�?”Tidy��_��"我知错了。。�?�? Windbreaker已经挂掉电话了。Tidy很苦��|��q�么��他真的会崩溃的�Q�聪明的读者，你能写个�E�序帮他完成�q�项工作吗？不过如果你的�E�序效率不够高的话，Tidy�q�是会受到Derek的责骂的.

Input

�W�一行一个整数T�Q�表�C�有T�l�数据�?br>每组数据�W�一行一个正整数N�Q�N<=50000�Q?表示敌�h有N个工兵营圎ͼ�接下来有N个正整数,�W�i个正整数ai代表�W�i个工兵营地里开始时有ai个�h�Q?<=ai<=50�Q��?br>接下来每行有一条命令，命��o�?�U��Ş式：
(1) Add i j,i和j为正整数,表示�W�i个营地增加j个�h�Q�j不超�q?0�Q?br>(2)Sub i j ,i和j为正整数,表示�W�i个营地减��j个�h�Q�j不超�q?0�Q?
(3)Query i j ,i和j为正整数,i<=j�Q�表�C��问第i到第j个营地的��M�h�?
(4)End 表示�l�束�Q�这条命令在每组数据最后出�?
每组数据最多有40000条命�?br>

Output

对第i�l�数�?首先输出“Case i:”和回�?
对于每个Query询问�Q�输��Z��个整数�ƈ回�R,表示询问的段中的��M�h�?�q�个数最多不��过1000000�?br>

Sample Input

1
10
1 2 3 4 5 6 7 8 9 10
Query 1 3
Add 3 6
Query 2 7
Sub 10 2
Add 6 3
Query 3 10
End

Sample Output

Case 1:
6
33
59

分析�Q�这题，最裸的�U�段树，最裸的树状数组。�ؓ了练习segment tree�Q�所�?..
直接��_��

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#define maxn 50010
using namespace std;
struct Tree
{
    int l, r, sum;
}tree[maxn * 5];
int p[maxn];
inline int L(int x)
{
    return (x << 1);
}
inline int R(int x)
{
    return (x << 1) | 1;
}
void build(int l, int r, int th)
{
    tree[th].l = l, tree[th].r = r, tree[th].sum = 0;
    if (l == r)
    {
        tree[th].sum = p[l];
        return;
    }
    int m = (l + r) >> 1;
    build(l, m, L(th)), build(m + 1, r, R(th));
    tree[th].sum = tree[L(th)].sum + tree[R(th)].sum;
}
void insert(int l, int r, int th, int num)
{
    if (tree[th].l == l && tree[th].r == r)
    {
        tree[th].sum += num;
        //if (tree[th].sum < 0) tree[th].sum = 0;有点多想�?wa了一��?nbsp;
        return;
    }
    int m = (tree[th].l + tree[th].r) >> 1;
    if (r <= m)
    {
        insert(l, r, L(th), num);
    }
    else if (l > m)
    {
        insert(l, r, R(th), num);
    }
    else
    {
        insert(l, m, L(th), num), insert(m + 1, r, R(th), num);
    }
    tree[th].sum = tree[L(th)].sum + tree[R(th)].sum;
}
int query(int l, int r, int th)
{
    if (tree[th].l == l && tree[th].r == r)
    {
        return tree[th].sum;
    }
    int m = (tree[th].l + tree[th].r) >> 1;
    if (r <= m)
    {
        return query(l, r, L(th));
    }
    else if (l > m)
    {
        return query(l, r, R(th));
    }
    else
    {
        return query(l, m, L(th)) + query(m + 1, r, R(th));
    }
}
int main()
{
    int t, n, i, a, b;
    char op[10];
    scanf("%d", &t);
    int ca = 0;
    while (t--)
    {
        ca++;
        printf("Case %d:\n", ca);
        scanf("%d", &n);
        for (i = 1; i <= n; i++)
        {
            scanf("%d", &p[i]);
        }
        build(1, n, 1);
        while (1)
        {
            scanf("%s", op);
            if (op[0] == 'E') break;
            scanf("%d%d", &a, &b);
            if (op[0] == 'Q')
            {
                int ans = query(a, b, 1);
                printf("%d\n", ans);
            }
            else if (op[0] == 'A')
            {
                insert(a, a, 1, b);
            }
            else
            {
                insert(a, a, 1, -b);
            }
        }

    }
    return 0;
}
/*
10
1 2 3 4 5 6 7 8 9 10
Q 1 1
Q 1 10

3
1 2 3
S 1 100
Q 1 1

*/

火碳�?/a> 2011-05-05 12:14 发表评论

Wed, 04 May 2011 05:54:00 GMT

Cube

Problem Description

Given an N*N*N cube A, whose elements are either 0 or 1. A[i, j, k] means the number in the i-th row , j-th column and k-th layer. Initially we have A[i, j, k] = 0 (1 <= i, j, k <= N).
We define two operations, 1: “Not” operation that we change the A[i, j, k]=!A[i, j, k]. that means we change A[i, j, k] from 0->1,or 1->0. (x1<=i<=x2,y1<=j<=y2,z1<=k<=z2).
0: “Query” operation we want to get the value of A[i, j, k].

Input

Multi-cases.
First line contains N and M, M lines follow indicating the operation below.
Each operation contains an X, the type of operation. 1: “Not” operation and 0: “Query” operation.
If X is 1, following x1, y1, z1, x2, y2, z2.
If X is 0, following x, y, z.

Output

For each query output A[x, y, z] in one line. (1<=n<=100 sum of m <=10000)

Sample Input

2 5
1 1 1 1  1 1 1
0 1 1 1
1 1 1 1  2 2 2
0 1 1 1
0 2 2 2

Sample Output

1
0
1

题意�Q�在一个三�l�长方体里把0元素�Ҏ��1或者把1�Ҏ��0或者询问某个位�|�是0�q�是1.
分析:
非树状数�l�莫属，其他数据�l�构都太帅了�Q�用不着�?br> 先说说一�l�上的树状数�l�，什么问题都得成最��单的开始，理解后才能推�q�到多维或复杂的�?br> 把对某个方体的修�Ҏ��作看做是�Ҏ��个方体的操作�?1�Q�那一个位�|�操作数的奇偶性决定了它的倹{�?br>树状数组1位置是管1的，2位置是管1�?�?3是管3的，4是管1�?�?.....所以对一个区间操作，�q�个区间首先按树状数�l�的分区规则�Q�会被分成若�q�个不相交的区间�Q�在树状数组里对应的��是若干不相交的��树�?br> 比如�?�?操作�Q�会分成分别对[1,2],[3,3]两段的操作；�?�?的操作，可以分成[2,2],[3,3],[4,4]�Q�但是对�q�种情况�Q�按�q�种方式��M��Ҏ��状数�l�是�ȝ��的，可以转化成分别对[1,4]操作�Q�再对[1,1]操作�Q�这样[1,1]被修改了两次�Q�所以相当于没修改过(配合�q�题的奇偶�?�Q�也可处理成对[1,4]的操作数+1�Q�再对[1,1]的操作数-1�?br> 知道每段的信息保存操作次敎ͼ�可以通过叠加��q��道每个位�|�的操作数�?br>比如�?�?操作了一�ơ，c[2] = 1(对应[1,2]的区�?�Q�c[3] = 1(对应[3,3]区间)�Q�再�?操作�Q�则c[1] = 1�Q�对应[1,1]区间�Q�。则1的操作数是c[1] + c[2]�?br> 到此�q�不理解的，��最好看�?a href="http://m.shnenglu.com/Ylemzy/articles/98322.html">树状数组的详�l�结�?/a>�Q�然后画一��d��应该清楚多了�Q�表达上的不便请原谅�Q�这个算是比较水的，不过�Ҏ��状数�l�的理解是有点用处的�Q�还有就是在树状数组上对方块的切�Ԍ��例如对对2�?的操作需要处理成[1,4]操作�Q�再对[1,1]操作 �Q�将�q�个操作推广�?�l��?br> 一�l�的是区间管区间�Q�二�l�是方块��理方块�Q�三�l�是方体��理方体�Q�本质跟一�l�一栗��?br>

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#define maxn 110
using namespace std;
int c[maxn][maxn][maxn];
int n;
void set()
{
    int i, j, k;
    for (i = 0; i <= n; i++)
    {
        for (j = 0; j <= n; j++)
        {
            for (k = 0; k <= n; k++)
            {
                c[i][j][k] = 0;
            }
        }
    }
}
inline int lowbit(int x)
{
    return x & (-x);
}
void update(int x, int y, int z)//分块更新或操�?nbsp;
{
    int i, j, k;
    for (i = x; i > 0; i -= lowbit(i))
    {
        for (j = y; j > 0; j -= lowbit(j))
        {
            for (k = z; k > 0; k -= lowbit(k))
            {
                c[i][j][k]++;
            }
        }
    }
}
int sum(int x, int y, int z)//向上叠加每个父节点的操作�ơ数
{
    int i, j, k, var = 0;
    for (i = x; i <= n; i += lowbit(i))
    {
        for (j = y; j <= n; j += lowbit(j))
        {
            for (k = z; k <= n; k += lowbit(k))
            {
                var += c[i][j][k];
            }
        }
    }
    return var;
}
int main()
{
    int m, op, x1, y1, z1, x2, y2, z2, ans;
    while (scanf("%d%d", &n, &m) - EOF)
    {
        set();
        while (m--)
        {
            scanf("%d%d%d%d", &op, &x1, &y1, &z1);
            if (op == 1)
            {
                scanf("%d%d%d", &x2, &y2, &z2);
                update(x2, y2, z2);
                update(x2, y1 - 1, z2);
                update(x1 - 1, y2, z2);
                update(x1 - 1, y1 - 1, z2);
                //--------
                update(x2, y2, z1 - 1);
                update(x2, y1 - 1, z1 - 1);
                update(x1 - 1, y2, z1 - 1);
                update(x1 - 1, y1 - 1, z1 - 1);
            }
            else
            {
                ans = sum(x1, y1, z1);
                printf("%d\n", ans % 2);
            }
        }
    }
    return 0;
}
/*
2 100
1 2 2 2 2 2 2
0 1 1 1

1
2 10
1 2 1 2 2
Q 2 2
1 2 1 2 1
Q 1 1
1 1 1 2 1
1 1 2 1 2
1 1 1 2 2
Q 1 1
1 1 1 2 1
Q 2 1
*/

火碳�?/a> 2011-05-04 13:54 发表评论

Sat, 23 Apr 2011 13:19:00 GMT

摘要: Marriage Match IV Problem Description Do not sincere non-interference。Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girl... 阅读全文

火碳�?/a> 2011-04-23 21:19 发表评论