【bzoj1176】[Balkan2007]Mokia

题目描述：

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.

输入：
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

输出：
对于每个输入2,输出一行,即输入2的答案

题解：

本蒟蒻的第一道cdq分治题。。。“cdq分治不就是归并排序吗？”写完这道题以后我对这句话有了更深的理解。

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#ifdef WIN32#define LL "%I64d"#else#define LL "%lld"#endif#ifdef CT#define debug(...) printf(__VA_ARGS__)#define setfile() #else#define debug(...)#define filename ""#define setfile() freopen(filename".in", "r", stdin); freopen(filename".out", "w", stdout);#endif#define R register#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T) ? EOF : *S++)#define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b))#define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b))#define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0)#define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0)char B[1 << 15], *S = B, *T = B;inline int FastIn(){R char ch; R int cnt = 0; R bool minus = 0;while (ch = getc(), (ch < '0' || ch > '9') && ch != '-') ;ch == '-' ? minus = 1 : cnt = ch - '0';while (ch = getc(), ch >= '0' && ch <= '9') cnt = cnt * 10 + ch - '0';return minus ? -cnt : cnt;}#define maxn 200010#define maxm 2000010struct event{int x, y, pos, opet, ans;inline bool operator < (const event &that) const {return pos < that.pos ;}}t[maxn], q[maxn];#define lowbit(_x) ((_x) & -(_x))int bit[maxm], last[maxm], s, w, cnt, now;inline void add(R int x, R int val){for (; x <= w; x += lowbit(x)){if (last[x] != now)bit[x] = 0;bit[x] += val;last[x] = now;}}inline int query(R int x){R int ans = 0;for (; x ; x -= lowbit(x)){if (last[x] == now)ans += bit[x];}return ans;}void cdq(R int left, R int right){if (left == right) return ;R int mid = left + right >> 1;cdq(left, mid); cdq(mid + 1, right);++now;for (R int i = left, j = mid + 1; j <= right; ++j){for (; i <= mid && q[i].x <= q[j].x; ++i)if (!q[i].opet)add(q[i].y, q[i].ans);if (q[j].opet)q[j].ans += query(q[j].y);}R int i, j, k = 0;for (i = left, j = mid + 1; i <= mid && j <= right; ){if (q[i].x <= q[j].x)t[k++] = q[i++];elset[k++] = q[j++];}for (; i <= mid; )t[k++] = q[i++];for (; j <= right; )t[k++] = q[j++];for (R int i = 0; i < k; ++i)q[left + i] = t[i];}int main(){//setfile();s = FastIn();w = FastIn();while (1){R int opt = FastIn();if (opt == 1){R int x = FastIn(), y = FastIn(), a = FastIn();q[++cnt] = (event){x, y, cnt, 0, a};}if (opt == 2){R int x = FastIn() - 1, y = FastIn() - 1, a = FastIn(), b = FastIn();q[++cnt] = (event) {x, y, cnt, 1, x * y * s};q[++cnt] = (event) {a, b, cnt, 2, a * b * s};q[++cnt] = (event) {x, b, cnt, 2, x * b * s};q[++cnt] = (event) {a, y, cnt, 2, a * y * s};}if (opt == 3) break;}cdq(1, cnt);std::sort(q + 1, q + cnt + 1);for (R int i = 1; i <= cnt; ++i)if (q[i].opet == 1)printf("%d\n",q[i].ans + q[i + 1].ans - q[i + 2].ans - q[i + 3].ans ), i += 3;return 0;}