数据结构训练
树状数组
题意：平面 \(n\) 个电子，初始速度分别为 \(\vec{v_i}\)，飙升系数 \(\sum_{1\le i\lt j\le n}|v_i\times v_j|^2\)，有两种操作:
- 1 p x y: 将 \(\vec{v_p}\) 改为 \((x, y)\)
- 2 l r: 询问 \([l,r]\) 这段区间内的电子的飘升系数
共 \(m\) 个，答案对 20170927 取模。
利用叉积公式，不妨设 \(v_i=(x_i,y_i)\)，则飙升系数为
\[\begin{array}{l}\displaystyle\sum_{1\le i\le j\le n}|v_i\times v_j|^2&=&\displaystyle\sum_{1\le i\lt j\le n}(x_iy_j-x_jy_i)^2\\
&=&\displaystyle\sum_{1\le i\lt j\le n}x_i^2y_j^2+\sum_{1\le i\lt j\le n}x_j^2y_i^2-2\sum_{1\le i\lt j\le n}(x_iy_i\cdot x_jy_j)\\
&=&\displaystyle\frac{1}{2}\sum_{1\le i,j\le n}x_i^2y_j^2+\frac{1}{2}\sum_{1\le i,j\le n}x_j^2y_i^2-\sum_{1\le i,j\le n}(x_iy_i\cdot x_jy_j)\\
&=&\displaystyle\sum_{1\le i\le n}x_i^2\sum_{1\le j\le n}y_i^2-(\sum_{1\le i\le n}x_iy_i)^2\end{array}\]
所以可以用树状数组维护 \(x_i,y_i,x_iy_i\) 的前缀和计算。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;
#define debug(x) cout << #x << " is " << x << endl
typedef long long ll;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f, N = 1e6 + 5, MOD = 20170927;
int n, m;
int a[N], b[N], c[N];
int u[N], v[N];
void update(int k, int x, int y) {
for (; k <= n; k += k & -k) {
a[k] = (a[k] + 1ll * x * x % MOD) % MOD;
b[k] = (b[k] + 1ll * y * y % MOD) % MOD;
c[k] = (c[k] + 1ll * x * y % MOD) % MOD;
}
}
void clear(int k, int x, int y) {
for (; k <= n; k += k & -k) {
a[k] = (a[k] - 1ll * x * x % MOD + MOD) % MOD;
b[k] = (b[k] - 1ll * y * y % MOD + MOD) % MOD;
c[k] = (c[k] - 1ll * x * y % MOD + MOD) % MOD;
}
}
int sum(int k, int *arr) {
int res = 0;
for (; k; k -= k & -k) (res += arr[k]) %= MOD;
return res;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%d%d", &u[i], &v[i]);
update(i, u[i], v[i]);
}
int op, p, x, y;
while (m--) {
scanf("%d", &op);
if (op == 1) {
scanf("%d%d%d", &p, &x, &y);
clear(p, u[p], v[p]);
update(p, x, y);
u[p] = x, v[p] = y;
}
else if (op == 2) {
scanf("%d%d", &x, &y);
int sa = (sum(y, a) - sum(x - 1, a) + MOD) % MOD;
int sb = (sum(y, b) - sum(x - 1, b) + MOD) % MOD;
int sc = (sum(y, c) - sum(x - 1, c) + MOD) % MOD;
printf("%lld\n", (1ll * sa * sb % MOD - 1ll * sc * sc % MOD + MOD) % MOD);
}
}
return 0;
}
RMQ 问题
题意：给出一个 \(n\) 行 \(m\) 列的矩阵，有 \(k\) 个问题，询问以 \(x_1\) 行 \(y_1\) 列为左上角，以 \(x_2\) 行 \(y_2\) 列为右下角的子矩阵的最大值。
考虑 \(f[i][j][p][q]\) 以 \((i,j)\) 为左上角，向下、向右延伸 \(2^p,2^q\) 的子矩阵的最大值，\(dp[i][j][0][0]\) 为本身。
- \(p\neq0,q=0\)，转化为一维 RMQ 有
\[f[i][j][p][0]=\max(f[i][j][p-1][0],f[i+2^{p-1}][j][p-1][0])
\]
- \(p=0,q\neq0\)，转化为一维 RMQ 有
\[f[i][j][0][q]=\max(f[i][j][0][q-1],f[i][j+2^{q-1}][0][q-1])
\]
- \(p\neq0,q\neq0\)，矩阵分成左右或上下两个部分，前面均处理过，直接取 \(\max\)
\[f[i][j][p][q]=\max(f[i][j][p-1][q],f[i+2^{p-1}][j][p-1][q])\\
or\quad f[i][j][p][q]=\max(f[i][j][p][q-1],f[i][j+2^{q-1}][p][q-1])\]
询问时从左上角、左下角、右上角、右下角转移过来，时间复杂度 \(O(nm\log(n)\log(m))\)。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;
#define debug(x) cout << #x << " is " << x << endl
typedef long long ll;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f, N = 255;
int n, m, k;
int lg[N], f[N][N][8][8];
int query(int ax, int ay, int bx, int by) {
int p = lg[bx-ax+1];
int q = lg[by-ay+1];
bx = bx - (1<<p) + 1;
by = by - (1<<q) + 1;
return max(max(f[ax][ay][p][q], f[ax][by][p][q]), max(f[bx][ay][p][q], f[bx][by][p][q]));
}
int main()
{
scanf("%d%d%d", &n, &m, &k);
lg[0] = -1;
for (int i = 1; i <= max(n, m); ++i) lg[i] = lg[i>>1] + 1;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
scanf("%d", &f[i][j][0][0]);
}
}
for (int p = 0; p <= lg[n]; ++p) {
for (int q = 0; q <= lg[m]; ++q) {
if (!(p + q)) continue;
for (int i = 1; i + (1<<p) - 1 <= n; ++i) {
for (int j = 1; j + (1<<q) - 1 <= m; ++j) {
if (p) f[i][j][p][q] = max(f[i][j][p-1][q], f[i+(1<<p-1)][j][p-1][q]);
else f[i][j][p][q] = max(f[i][j][p][q-1], f[i][j+(1<<q-1)][p][q-1]);
}
}
}
}
int ax, ay, bx, by;
while (k--) {
scanf("%d%d%d%d", &ax, &ay, &bx, &by);
printf("%d\n", query(ax, ay, bx, by));
}
return 0;
}
线段树
posted by 2inf
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