【LG1527】[国家集训队]矩阵乘法

【LG1527】[国家集训队]矩阵乘法

题面

洛谷

题解

我也不知道为什么取个这样的名字。。。

其实就是区间\(kth\)扩展到二维

还是用整体二分搞啦，把树状数组换成二维的

其他的基本没有什么差别

复杂度\(nlog^3\)

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
namespace IO {
    const int BUFSIZE = 1 << 20;
    char ibuf[BUFSIZE], *is = ibuf, *it = ibuf;
    inline char gc() {
        if (is == it) it = (is = ibuf) + fread(ibuf, 1, BUFSIZE, stdin);
		return *is++;
    }
}
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (ch != '-' && (ch > '9' || ch < '0')) ch = IO::gc();
    if (ch == '-') w = -1 , ch = IO::gc();
    while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = IO::gc();
    return w * data;
}
const int SIZE = 400005, MAX_N = 505;
struct Query { int op, x, y, _x, _y, k; } q[SIZE], rq[SIZE], lq[SIZE];
void Output(int x) {
    printf("%d %d %d %d %d %d\n", q[x].op, q[x].x, q[x].y, q[x]._x, q[x]._y, q[x].k);
}
int N, M, ans[SIZE];
int c[MAX_N][MAX_N];
inline int lb(int x) { return x & -x; }
void add(int x, int y, int v) {
    for (int i = x; i <= N; i += lb(i))
        for (int j = y; j <= N; j += lb(j))
            c[i][j] += v;
}
int sum(int x, int y) {
	int res = 0;
	for (int i = x; i > 0; i -= lb(i))
	    for (int j = y; j > 0; j -= lb(j))
	        res += c[i][j];
	return res;
}
void Div (int lval, int rval, int st, int ed) {
    if (st > ed) return ;
    if (lval == rval) {
        for (int i = st; i <= ed; i++) if (q[i].op != 0) ans[q[i].op] = lval;
        return ;
    }
    int mid = (lval + rval) >> 1;
    int lt = 0, rt = 0;
    for (int i = st; i <= ed; i++) {
        int x = q[i].x, y = q[i].y;
        if (q[i].op == 0) {
            if (q[i].k <= mid) lq[++lt] = q[i], add(x, y, 1);
            else rq[++rt] = q[i];
        } else {
            int _x = q[i]._x, _y = q[i]._y;
            int res = sum(_x, _y) - sum(x - 1, _y) - sum(_x, y - 1) + sum(x - 1, y - 1);
            if (q[i].k <= res) lq[++lt] = q[i];
            else q[i].k -= res, rq[++rt] = q[i];
        }
    }
    for (int i = st; i <= ed; i++) if (q[i].op == 0 && q[i].k <= mid) add(q[i].x, q[i].y, -1);
    for (int i = 1; i <= lt; i++) q[st + i - 1] = lq[i];
    for (int i = 1; i <= rt; i++) q[st + lt + i - 1] = rq[i];
    Div(lval, mid, st, st + lt - 1);
    Div(mid + 1, rval, st + lt, ed);
}
int main () {
    N = gi(), M = gi();
    int cnt = 0;
    for (int i = 1; i <= N; i++)
        for (int j = 1; j <= N; j++)
            q[++cnt] = (Query){0, i, j, 0, 0, gi()};
    for (int i = 1; i <= M; i++) q[++cnt] = (Query){i, gi(), gi(), gi(), gi(), gi()};
    Div(1, 1e9, 1, cnt);
    for (int i = 1; i <= M; i++) printf("%d\n", ans[i]);
    return 0;
}