BZOJ 1176: [Balkan2007]Mokia( CDQ分治 + 树状数组 )

考虑cdq分治, 对于[l, r)递归[l, m), [m, r); 然后计算[l, m)的操作对[m, r)中询问的影响就可以了. 具体就是差分答案+排序+离散化然后树状数组维护.操作数为M的话时间复杂度大概是O(M(logM)^2)

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cctype>

 

using namespace std;

 

typedef long long ll;

#define cal(x, y) (ll(Val) * (x) * (y))

#define lowbit(x) ((x) & -(x))

#define h(v) (lower_bound(Y, Y + Yn, v) - Y + 1)

#define Q(x) o[Que[x]]

#define A(x) o[Ad[x]]

 

const int MAXN = 640009;

const int MAXQ = 40009;

 

inline int read() {

char c = getchar();

int ret = 0, t = 0;

for(; !isdigit(c); c = getchar()) if(c == '-') t = 1;

for(; isdigit(c); c = getchar()) ret = ret * 10 + c - '0';

return t ? -ret : ret;

}

 

int Val, On, Qn;

int ans[MAXQ];

int Ad[MAXN], Que[MAXQ], Adn, Quen;

int Y[MAXN], Yn, X[MAXN], Xn;

 

struct O {

int p, x, y, v;

inline void Set(int _p, int _x, int _y, int _v) {

p = _p; x = _x; y = _y; v = _v;

}

bool operator < (const O &o) const {

return x < o.x;

}

} o[MAXN + MAXQ];

 

void Init() {

Val = read();

Qn = On = 0;

for(int t = read(); (t = read()) != 3; ) {

if(t == 1) {

int x = read(), y = read();

o[On++].Set(-1, x, y, read());

} else {

int x0 = read() - 1, y0 = read() - 1, x1 = read(), y1 = read();

o[On++].Set(Qn, x0, y0, 1);

o[On++].Set(Qn, x1, y1, 1);

o[On++].Set(Qn, x0, y1, 0);

o[On++].Set(Qn, x1, y0, 0);

ans[Qn++] = cal(x0, y0) + cal(x1, y1) - cal(x1, y0) - cal(x0, y1);

}

}

}

 

struct BIT {

int b[MAXN];

BIT() {

memset(b, 0, sizeof b);

}

void Add(int p, int v) {

for(; p <= Yn; p += lowbit(p)) b[p] += v;

}

int Query(int p) {

int ret = 0;

for(; p; p -= lowbit(p)) ret += b[p];

return ret;

}

} Bit;

 

bool Cmp(const int &l, const int &r) {

return o[l].x < o[r].x;

}

 

void Solve() {

Yn = Xn = 0;

for(int i = 0; i < Quen; i++)

Y[Yn++] = Q(i).y, X[Xn++] = Q(i).x;

for(int i = 0; i < Adn; i++)

Y[Yn++] = A(i).y, X[Xn++] = A(i).x;

sort(Que, Que + Quen, Cmp);

sort(Ad, Ad + Adn, Cmp);

sort(Y, Y + Yn); Yn = unique(Y, Y + Yn) - Y;

sort(X, X + Xn); Xn = unique(X, X + Xn) - X;

int _Adn = 0, _Quen = 0;

for(int i = 0; i < Xn; i++) {

while(_Adn < Adn && A(_Adn).x == X[i])

Bit.Add(h(A(_Adn).y), A(_Adn).v), _Adn++;

while(_Quen < Quen && Q(_Quen).x == X[i]) {

int ret = Bit.Query(h(Q(_Quen).y));

ans[Q(_Quen).p] += Q(_Quen).v ? ret : -ret;

_Quen++;

}

}

for(int i = 0; i < Adn; i++)

Bit.Add(h(A(i).y), -A(i).v);

}

 

// [l, r)

void CDQ(int l, int r) {

if(l + 1 >= r) return;

int m = (l + r) >> 1;

CDQ(l, m); CDQ(m, r);

Adn = Quen = 0;

for(; l < m; l++)

if(!~o[l].p) Ad[Adn++] = l;

for(; l < r; l++)

if(~o[l].p) Que[Quen++] = l;

if(Quen && Adn) Solve();

}

 

int main() {

Init();

CDQ(0, On);

for(int i = 0; i < Qn; i++)

printf("%d\n", ans[i]);

return 0;

}

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1176: [Balkan2007]Mokia

Time Limit: 30 Sec  Memory Limit: 162 MB
Submit: 1281  Solved: 537
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Description

维护一个W*W的矩阵，初始值均为S.每次操作可以增加某格子的权值,或询问某子矩阵的总权值.修改操作数M<=160000,询问数Q<=10000,W<=2000000.

Input

第一行两个整数,S,W;其中S为矩阵初始值;W为矩阵大小

接下来每行为一下三种输入之一(不包含引号):

"1 x y a"

"2 x1 y1 x2 y2"

"3"

输入1:你需要把(x,y)(第x行第y列)的格子权值增加a

输入2:你需要求出以左上角为(x1,y1),右下角为(x2,y2)的矩阵内所有格子的权值和,并输出

输入3:表示输入结束

Output

对于每个输入2,输出一行,即输入2的答案

Sample Input

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

Sample Output

3
5

HINT

保证答案不会超过int范围

Source