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{//leftpushdown(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];}}//rightpushdown(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;
}

转载于:https://www.cnblogs.com/Blackops/p/7223649.html