hdu 3854 Glorious Array

Problem Description

You are given a array with N nodes. The array nodes are numbered from 1 to N ( represented by a_i ). In the start, the color
of any node in the array is white or black. We define that the "distance" between a and b is min{a_i| a<=i<=b} if a-th and bth
nodes have diffrerent color, or infinite if they have the same color.
We will ask you to perfrom some instructions of the following form:
0 i : change the color of the i-th node (from white to black, or from black to white);
or
1 : ask for the kinds of diffrent pair of nodes' distance is less than K

 

Input

The first line contains a positive integer: the number of test cases, at most 100.
After that per test case:
One line with three integers N, m and K (1 ≤ N, m ≤ 1 00000, K ≤ 1000000): the length of the array, the number of
instructions and K which used in operation 1 .
Then comes a line with N intergers a_i (|a_i|<10000000).
A line with N intergers c_i (c_i = 0(white) or 1(black)), which reprsent the color of N nodes.
Next m lines, each will be a instruction described above.

 

Output

one interger for each operation 1.

 

Sample Input

1 5 3 2 2 3 1 4 2 0 1 0 1 1 1 0 2 1

 

Sample Output

5 6

//

题目描述：
给你n个数，和每个数字的状态（0或者1）。
有两种操作：
第一种：翻转某个数字的状态。
第二种：询问，要求输入满足条件的i， j组合个数。
一个满足条件的i，j要求：
A：1 <= i <= j <= n
B: i和j数字的状态不同
C：i和j的闭区间内的数字的最小值小于k（k由输入给定）
解题报告：
对于一开始的输入先求出组合个数。
然后对于每次的翻转状态，更新组合个数：
比如翻转i，如果x[i] < k, 那么只需看i左侧有多少个0或者1（具体看x[i]当前状态），和i右侧有多少个0或者1（同上），就可以知道这次翻转具体变化了多少，具体的实现可以用树状数组。
如果x[i] > k，初始化时记下在i左侧最近的小于k的数字的位置L[i]，同理R[i]。
找到L[i]左侧0或者1的个数和R[i]右侧0或者1的个数，就是这次变化的个数。
应该比较好理解。
代码如下：

#include

#include

#include

using namespace std;

#define typev int // type of res

#define N 200000

typev ar[N]; // index: 1 ~ N

int lowb(int t) { return t & (-t) ; }

void add(int i, typev v)

{for ( ; i < N; ar[i] += v, i += lowb(i));}

typev sum(int i)

{

    typev s = 0;

    for ( ; i > 0; s += ar[i], i -= lowb(i));

    return s;

}

int n, m, k;

long long nowans;

int x[N], L[N], R[N], sta[N];

void init()

{

    memset(L, -1, sizeof(L)); memset(R, -1, sizeof(R));

    memset(ar, 0, sizeof(ar));

    for(int i = 1; i <= n; i++) if (x[i] < k)

        for(int j = i + 1; j <= n && x[j] >= k; j++)

            L[j] = i;

    for(int i = n; i >= 1; i--) if (x[i] < k)

        for(int j = i - 1; j >= 1 && x[j] >= k; j--)

            R[j] = i;

}

int Get(int l, int r, int now)

{

    //printf("l %d r %d %d %dn", l, r, sum(r), sum(l - 1));

    if (l > r) return 0;

    if (l == -1 || r == -1) return 0;

    if (now == 0) return r - l + 1 - (sum(r) - sum(l - 1));

    return sum(r) - sum(l - 1);

}

int main()

{

    int t;scanf("%d", &t);

    while(t--)

    {

        scanf("%d%d%d", &n, &m, &k);

        for(int i = 1; i <= n; i++) scanf("%d", &x[i]);

        init();

        for(int i = 1; i <= n; i++)

        {

            scanf("%d", &sta[i]);

            if (sta[i]) add(i, 1);

        }

        nowans = 0;

        for(int i = 1; i <= n; i++)

            if (x[i] < k) nowans += Get(i + 1, n, !sta[i]);

            else nowans += Get(R[i], n, !sta[i]);

        //cout << nowans << endl;

        while(m--)

        {

            int tmp;scanf("%d", &tmp);

            if (tmp == 1) cout << nowans << endl;

            else

            {

                scanf("%d", &tmp);

                int before = 0, after = 0;

                int i = tmp;

                if (x[tmp] < k)

                {

                    before += Get(1, i - 1, !sta[i]) + Get(i + 1, n, !sta[i]);

                    after += Get(1, i - 1, sta[i]) + Get(i + 1, n, sta[i]);

                }else

                {

                    before += Get(1, L[i], !sta[i]) + Get(R[i], n, !sta[i]);

                    after += Get(1, L[i], sta[i]) + Get(R[i], n, sta[i]);

                }

                if (sta[i] == 0) add(i, 1), sta[i] = 1;

                else add(i, -1), sta[i] = 0;

                nowans += (after - before);

            }

        }

    }

    return 0;

}