NBUT校赛 J Alex’s Foolish Function（分块+延迟标记）

Problem J: Alex’s Foolish Function

Time Limit: 8 Sec Memory Limit: 128 MB
Submit: 18 Solved: 2

Description

Alex has an array F which size is n, initially each element’s value is zero. Now he wants to operate the array, he has 4 different functions:

1) Function A(l, r):
Add the elements between l to r (inclusive) where the ith add i - l + 1, for instance, if l is 4,

r is 6 and the elements between 4 to 6 is 3 4 10, after this operation, these elements will become 4 6 13.

2) Function B(l, r):
Add the elements between l to r (inclusive) where the ith add r - i + 1, for instance, if l is 4,

r is 6 and the elements between 4 to 6 is 3 4 10, after this operation, these elements will become 6 6 11.

3) Function C(l, r, x):
Set all the elements(between l to r) to x, for instance, if l is 4, r is 6 and the elements

between 4 to 6 is 3 4 10, and x = 2, after this operation, these elements will become 2 2 2.

4) Function S(l, r):
Output Fl + Fl+1 + Fl+2 + ...+ Fr.

Input

Input start with an integer T(1 <= T <= 5), denoting the number of test case.

For each test case, first line contains n, m (1 <= n <= 200000, 1 <= m <= 100000), denoting the array’size and the number of operations.

Next m lines, each line contains an operation, formatting as

A l r
B l r
C l r x

S l r

Output

For each S Function operations, output the sum.

Sample Input

Sample Output

正规做法是线段树维护区间的左加、右加或者覆盖信息，然而想用分块写一写。

对于A操作，每一个块维护起始的加数，以及A操作的次数，

对于B操作，每一个块维护末尾的加数，以及B操作的次数，

对于C操作，每一个块维护准备覆盖的值val。

显然对于A操作和B操作维护的信息，是满足区间加法的，即[1,4]A操作一次，[2,4]A操作一次，比如当前块是[2,4]，记为Bx，那么我们可以记录起始的加数：(2-1+1)+(2-2+1)=3，A操作次数为2，即arr[2]+=3，arr[3]+=(3+2)，arr[4]+=(3+2+2)，然后我们可以发现这是一个等差数列，可以用公式快速得到有延迟标记A操作的区间和，即$(Bx.lazy_A+(Bx.lazy_A+Bx.len-1))*Bx.len/2+Bx.sum$，然后B操作也同理，最后加上本身的$Bx.sum$即可（这里要注意AB操作不能重复加Bx.sum）；B操作同理；C操作的话由于是覆盖那么在更新的时候把更新到的块的$lazy_c$更新，并把$lazy_A$与$lazy_B$取消掉即可，由于可能$lazy_c$为0，初始值应设为一个不太可能用到的数，比如$-INF$。

代码：

#include <bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

#define LC(x) (x<<1)

#define RC(x) ((x<<1)+1)

#define MID(x,y) ((x+y)>>1)

#define fin(name) freopen(name,"r",stdin)

#define fout(name) freopen(name,"w",stdout)

#define CLR(arr,val) memset(arr,val,sizeof(arr))

#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);

typedef pair<int, int> pii;

typedef long long LL;

const double PI = acos(-1.0);

const int N = 200010;

const int M = 450;

struct Block

{

    int l, r;

    LL lazy1, lazy2, lazy3, sum;

    LL c1, c2;

    inline int len()

    {

        return r - l + 1;

    }

};

Block B[M];

LL arr[N];

int belong[N], unit, bcnt;

int n, m;

void init()

{

    int i;

    unit = sqrt(n);

    bcnt = n / unit;

    if (n % unit)

        ++bcnt;

    for (i = 1; i <= bcnt; ++i)

    {

        B[i].l = (i - 1) * unit + 1;

        B[i].r = i * unit;

        B[i].sum = 0LL;

        B[i].lazy1 = 0LL;

        B[i].lazy2 = 0LL;

        B[i].lazy3 = -INF;

        B[i].c1 = B[i].c2 = 0LL;

    }

    B[bcnt].r = n;

    for (i = 1; i <= n; ++i)

    {

        arr[i] = 0LL;

        belong[i] = (i - 1) / unit + 1;

    }

}

void pushdown(int bx)

{

    if (B[bx].lazy3 != -INF)

    {

        for (int i = B[bx].l; i <= B[bx].r; ++i)

            arr[i] = B[bx].lazy3;

        B[bx].lazy3 = -INF;

    }

    if (B[bx].lazy1)

    {

        for (int i = B[bx].l; i <= B[bx].r; ++i)

        {

            arr[i] += + B[bx].lazy1;

            B[bx].lazy1 += B[bx].c1;

        }

        B[bx].lazy1 = 0LL;

        B[bx].c1 = 0LL;

    }

    if (B[bx].lazy2)

    {

        for (int i = B[bx].r; i >= B[bx].l; --i)

        {

            arr[i] += + B[bx].lazy2;

            B[bx].lazy2 += B[bx].c2;

        }

        B[bx].lazy2 = 0LL;

        B[bx].c2 = 0LL;

    }

    B[bx].sum = 0LL;

    for (int i = B[bx].l; i <= B[bx].r; ++i)

        B[bx].sum += arr[i];

}

void update(int l, int r, int ops, LL val = 0LL)

{

    int bl = belong[l], br = belong[r];

    int i;

    if (bl == br)

    {

        pushdown(bl);

        if (ops == 1)

        {

            for (i = l; i <= r; ++i)

            {

                B[bl].sum -= arr[i];

                arr[i] = arr[i] + (LL)(i - l + 1LL);

                B[bl].sum += arr[i];

            }

        }

        else if (ops == 2)

        {

            for (i = r; i >= l; --i)

            {

                B[bl].sum -= arr[i];

                arr[i] = arr[i] + (LL)(r - i + 1LL);

                B[bl].sum += arr[i];

            }

        }

        else if (ops == 3)

        {

            for (i = l; i <= r; ++i)

            {

                B[bl].sum -= arr[i];

                arr[i] = val;

                B[bl].sum += arr[i];

            }

        }

    }

    else

    {

        //left

        pushdown(bl);

        if (ops == 1)

        {

            for (i = l; i <= B[bl].r; ++i)

            {

                B[bl].sum -= arr[i];

                arr[i] = arr[i] + (LL)(i - l + 1LL);

                B[bl].sum += arr[i];

            }

        }

        else if (ops == 2)

        {

            for (i = B[bl].r; i >= l; --i)

            {

                B[bl].sum -= arr[i];

                arr[i] = arr[i] + (LL)(r - i + 1LL);

                B[bl].sum += arr[i];

            }

        }

        else if (ops == 3)

        {

            for (i = l; i <= B[bl].r; ++i)

            {

                B[bl].sum -= arr[i];

                arr[i] = val;

                B[bl].sum += arr[i];

            }

        }

        //right

        pushdown(br);

        if (ops == 1)

        {

            for (i = B[br].l; i <= r; ++i)

            {

                B[br].sum -= arr[i];

                arr[i] = arr[i] + (LL)(i - l + 1LL);

                B[br].sum += arr[i];

            }

        }

        else if (ops == 2)

        {

            for (i = r; i >= B[br].l; --i)

            {

                B[br].sum -= arr[i];

                arr[i] = arr[i] + (LL)(r - i + 1LL);

                B[br].sum += arr[i];

            }

        }

        else if (ops == 3)

        {

            for (i = B[br].l; i <= r; ++i)

            {

                B[br].sum -= arr[i];

                arr[i] = val;

                B[br].sum += arr[i];

            }

        }

        for (i = bl + 1; i < br; ++i)

        {

            if (ops == 3)

            {

                B[i].lazy3 = val;

                B[i].c1 = 0LL;

                B[i].c2 = 0LL;

                B[i].lazy1 = 0LL;

                B[i].lazy2 = 0LL;

            }

            else if (ops == 2)

            {

                B[i].lazy2 += (LL)(r - B[i].r + 1LL);

                ++B[i].c2;

            }

            else if (ops == 1)

            {

                B[i].lazy1 += (LL)(B[i].l - l + 1LL);

                ++B[i].c1;

            }

        }

    }

}

LL query(int l, int r)

{

    int bl = belong[l], br = belong[r], i;

    LL sum = 0LL;

    if (bl == br)

    {

        pushdown(bl);

        for (i = l; i <= r; ++i)

            sum += arr[i];

        return sum;

    }

    else

    {

        pushdown(bl);

        pushdown(br);

        for (i = l; i <= B[bl].r; ++i)

            sum += arr[i];

        for (i = B[br].l; i <= r; ++i)

            sum += arr[i];

        for (i = bl + 1; i < br; ++i)

        {

            if (B[i].lazy3 != -INF)

                sum += (LL)B[i].len() * B[i].lazy3;

            if (B[i].lazy1)

                sum += (B[i].lazy1 + B[i].lazy1 + (LL)(B[i].len() - 1LL) * B[i].c1) * (LL)B[i].len() / 2LL;

            if (B[i].lazy2)

                sum += (B[i].lazy2 + B[i].lazy2 + (LL)(B[i].len() - 1LL) * B[i].c2) * (LL)B[i].len() / 2LL;

            if (B[i].lazy3 == -INF)

                sum += B[i].sum;

        }

        return sum;

    }

}

int main(void)

{

    // fin("test0.in");

    // fout("data_wa.txt");

    int tcase, l, r, v;

    char ops[4];

    scanf("%d", &tcase);

    while (tcase--)

    {

        scanf("%d%d", &n, &m);

        init();

        while (m--)

        {

            scanf("%s", ops);

            if (ops[0] == 'A')

            {

                scanf("%d%d", &l, &r);

                update(l, r, 1);

            }

            else if (ops[0] == 'B')

            {

                scanf("%d%d", &l, &r);

                update(l, r, 2);

            }

            else if (ops[0] == 'C')

            {

                scanf("%d%d%d", &l, &r, &v);

                update(l, r, 3, (LL)v);

            }

            else

            {

                scanf("%d%d", &l, &r);

                printf("%lld\n", query(l, r));

            }

        }

    }

    return 0;

}