Codeforces 390E Inna and Large Sweet Matrix 树状数组改段求段

题目链接：点击打开链接

题意：给定n*m的二维平面 w个操作

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 = 纵坐标在 y1,y2间的总数

ans2 = 横坐标不在x1,x2间的总数

puts(ans1-ans2);

more format:

for i : 1 to n

for j : y1 to y2

ans1 += mp[i][j]

因为n最大是4e6, 所以用树状数组改段求段代替线段树

#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>template <class T>inline bool rd(T &ret) {char c; int sgn;if (c = getchar(), c == EOF) return 0;while (c != '-' && (c<'0' || c>'9')) c = getchar();sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');ret *= sgn;return 1;}template <class T>inline void pt(T x) {if (x <0) {putchar('-');x = -x;}if (x>9) pt(x / 10);putchar(x % 10 + '0');}using namespace std;typedef long long ll;const int N = 4e6 + 100;template<class T>struct Tree{T 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; }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){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){return get_pre(r) - get_pre(l - 1);}};Tree<ll> x, y;int main(){int n, m, w;rd(n); rd(m); rd(w);x.init(n); y.init(m);ll all = 0;while (w--){int op, x1, x2, y1, y2; ll value;rd(op); rd(x1); rd(y1); rd(x2); rd(y2);if (op == 0){rd(value);all += value * (x2 - x1 + 1) * (y2 - y1 + 1);x.add(x1, x2, value * (y2 - y1 + 1));y.add(y1, y2, value * (x2 - x1 + 1));}else{pt(y.get(1, y2) - y.get(1, y1 - 1) - (all - x.get(1, x2) + x.get(1, x1 - 1))); puts("");}}return 0;}