Codeforces 390E Inna and Large Sweet Matrix (tree array change segment)

The tree array can only implement the function of section modification and interval query, instead of the line tree without lazy tag, and the code quantity and constant are small .

First define an array of int c[n]; and empty memset (c, 0, sizeof C);

1, Single point modification: c[x] + = y; The corresponding function is the change (x, y);

2. Prefix and: the corresponding function is int sum (x)

The complexity of both operations is O (LOGN)

The template is as follows:

int c[n], MAXN;  inline int lowbit (int x) {return x& (-X);}  void change (int i, int x)//i point increment is x  {while      (I <= maxn)      {          C[i] + = x;          i + = Lowbit (i);      }  }  int sum (int x) {//Interval sum [1,x]      int ans = 0;      for (int i = x; I >= 1; I-= Lowbit (i))          ans + = c[i];      return ans;  }  

How to use a tree-like array for interval operation

First, define two tree arrays X, Y

Now we need an array of int a[n]; Interval operation: [L, R] + = val is for i:l to R a[i] + = val;

Then define an int size = r-l+1, which is the interval length

The corresponding modification is

1, x[l] + = val; X[r+1]-= Val;

2, y[l] + =-1 * val * (L-1); Y[r+1] + = val * R;

The corresponding query is

When we sum the operation in the tree array is ans = x.sum (k) * k + y.sum (k)

Category discussion k respectively in [1,l-1], [L, R], [r+1, +]

1, K[1,l-1]

Obviously x.sum (k) = = 0 and y.sum (k) = = 0-ans = x.sum (k) *k + y.sum (k) = 0*i+0 = 0 The result is in accordance with the actual.

2, K[l, R]

X.sum (k) * k = x[l] * k = val * k, y.sum (k) = y[l] = 1 * val * (L-1)

Ans = val * k-val * (L-1) = Val * (k-(L-1));

3, K[r+1,]

X.sum (k) * k = (X[l] + x[r]) * k = 0 * k = 0;

Y.sum (k) = Y[l] + y[r] =-val * (L-1) + val * R = val * (r-l+1) = val * size

X.sum (k) * k + y.sum (k) = val * size

Proof is complete.

The two tree arrays in the following template c[0], c[1] correspond to the above X, Y

Interval modification: Add (L, R, Val)

Prefix and Get_pre (R) for int a[n]

Interval query: Get (L,R)

const int N = 4e5 + 100;template<class t>struct tree{t c[2][n];int maxn;void init (int x) {maxn = x+10; memset (c, 0, S Izeof c);} inline int lowbit (int x) {return x&-x;} T sum (t *b, int x) {T ans = 0;if (x = = 0) ans = b[0];while (x) ans + = b[x], X-= Lowbit (x); return ans; void Change (t *b, int x, t value) {if (x = = 0) b[x] + = value, X++;while (x <= maxn) b[x] + = value, x + = Lowbit (x);} T get_pre (int r) {return sum (c[0], R) * r + sum (c[1], r);} void Add (int l, int r, T value) {//Interval weighted change (c[0], L, value); Change (c[0], R + 1,-value); Change (c[1], L, Value * (-L + 1)) ; Change (c[1], R + 1, value * r);} T get (int l, int r) {//Interval sum return Get_pre (R)-Get_pre (L-1);}}; Tree<ll> Tree;

Well, back to the point, let's talk about the test instructions of the problem:

Test instructions: A two-dimensional planar w operation for a given n*m

int Mp[n][m] = {0};

1, 0 (x1,y1) (x2,y2) value

For i:x1 to X2

For J:y1 to Y2

MP[I][J] + = value;

2, 1 (x1, y1) (x2 y2)

ANS1 = total number of ordinate between y1,y2

ANS2 = The total number of horizontal axis not x1,x2 between

Puts (ANS1-ANS2);

The code is as follows:

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm>using namespace      Std;typedef Long long ll;const int n = 4e6+10;int n, m, w;template<class t> struct tree{ll c[2][n];      int MAXN;      void init (int x) {maxn = x+10; memset (c, 0, sizeof C);      } inline int lowbit (int x) {return x&-x;}          ll sum (ll *b, int x) {ll ans = 0;          if (x = = 0) ans = b[0];          while (x) ans + = b[x], X-= Lowbit (x);      return ans;          } void Change (ll *b, int x, ll value) {if (x = = 0) b[x] + = value, x + +;      while (x <= maxn) b[x] + = value, x + = Lowbit (x);      } ll Get_pre (int r) {return sum (c[0], R) * r + sum (c[1], R);          } void Add (int l, int r, ll value) {change (c[0], L, value);          Change (C[0], R + 1,-value);          Change (c[1], L, Value * (-L + 1));      Change (C[1], R + 1, value * r); } ll get (int l, int r) {return Get_pre(r)-Get_pre (L-1);  }  };  Tree<ll> x, y; int main () {scanf ("%d%d%d", &n, &m, &w); X.init (n); Y.init (m); int tmp;ll all = 0;while (w--) {scanf ("%d", &t MP); int x1, x2, y1, y2, v;if (tmp = = 0) {scanf ("%d%d%d%d%d", &x1, &y1, &x2, &y2, &v); all + = v * (x2-x1+1 ) * (y2-y1+1); X.add (x1, x2, V * (y2-y1 + 1)); Y.add (y1, y2, V * (x2-x1 + 1));} if (tmp = = 1) {scanf ("%d%d%d%d", &x1, &y1, &x2, &y2);p rintf ("%i64d\n", Y.get (1, y2)-y.get (1, y1-1)-(All -X.get (1, x2) + x.get (1, x1-1));}} return 0;}

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Codeforces 390E Inna and Large Sweet Matrix (tree array change segment)