[Educational Codeforces Round 56 (Rated for Div. 2)] —G. Multidimensional Queries（二进制状压，线段树）
G. Multidimensional Queries
time limit per test
6 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output
You are given an array aa of nn points in kk-dimensional space. Let the distance between two points axax and ayay be ∑i=1k|ax,i−ay,i|∑i=1k|ax,i−ay,i| (it is also known as Manhattan distance).
You have to process qq queries of the following two types:
- 11 ii b1b1 b2b2 ... bkbk — set ii-th element of aa to the point (b1,b2,…,bk)(b1,b2,…,bk);
- 22 ll rr — find the maximum distance between two points aiai and ajaj, where l≤i,j≤rl≤i,j≤r.
题意：
在一个(mathit k)维度坐标系中，点(a_x,a_y)的距离为(sum limits_{i = 1}^{k} |a_{x, i} - a_{y, i}|)。
现在给定(mathit n)个点，和(mathit q)个询问，每一个询问有( ext 2)种：
- $1 i b_1 b_2 ... b_k (代表将第)mathit i(个点的坐标改为)b_1 b_2 ... b_k$
- (2 l r)，代表让你找出两个距离最大的节点(a_i,a_j,l le i, j le r)，输出他们的距离即可。
思路：
点(a_x,a_y)的距离为(sum limits_{i = 1}^{k} |a_{x, i} - a_{y, i}|=max(sum_{i=1}^{k}Q_i*a_i+sum_{i=1}^{k}(!Q_i)*b_i,Q_iin{{-1,1}}))。
仔细思考上面等式，右边表达式中会有很多无效的状态，但是所有状态中的最大值一定是等于左侧表达式的。
然后我们可以建立(2^k)个线段树，第(mathit i)颗数维护的是((i)_2)中第(mathit j)位为( ext 1)时(a_{i,j})取正好，否则取符号的总和最大值。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <unordered_map>
// #include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '
' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '
' : ' ');}}
const int maxn = 200010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
ll segment_tree[maxn << 2][35];
int n, k;
int a[maxn][6];
void pushup(int rt)
{
repd(j, 0, (1<<k)- 1)
{
segment_tree[rt][j] = max(segment_tree[rt << 1][j], segment_tree[rt << 1 | 1][j]);
}
}
void show(int rt)
{
repd(j, 0, k - 1)
printf("%d %d %lld
",rt,j,segment_tree[rt][j]);
}
void build(int rt, int l, int r)
{
if (l == r)
{
for (int i = 0; i < (1 << k); ++i)
{
ll sum = 0ll;
repd(j, 0, k - 1)
{
if (i & (1 << j))
{
sum += a[l][j + 1];
} else
{
sum -= a[l][j + 1];
}
}
// printf("%d %d %lld
",l,i,sum);
segment_tree[rt][i] = sum;
}
} else
{
int mid = (l + r) >> 1;
build(rt << 1, l, mid);
build(rt << 1 | 1, mid + 1, r);
pushup(rt);
}
// show(rt);
}
void update(int rt, int l, int r, int id)
{
if (l == r && l == id)
{
for (int i = 0; i < (1 << k); ++i)
{
ll sum = 0ll;
repd(j, 0, k - 1)
{
if (i & (1 << j))
{
sum += a[id][j + 1];
} else
{
sum -= a[id][j + 1];
}
}
segment_tree[rt][i] = sum;
}
return ;
} else
{
int mid = (l + r) >> 1;
if (id > mid)
update(rt << 1 | 1, mid + 1, r, id);
else
update(rt << 1, l, mid, id);
pushup(rt);
}
}
ll f[35];
void query(int rt, int l, int r, int ql, int qr)
{
if (l >= ql && qr >= r)
{
for (int i = 0; i < (1 << k); ++i)
{
f[i] = max(f[i], segment_tree[rt][i]);
}
} else
{
int mid = (l + r) >> 1;
if (ql <= mid)
{
query(rt << 1, l, mid, ql, qr);
}
if (qr > mid)
{
query(rt << 1 | 1, mid + 1, r, ql, qr);
}
}
}
int main()
{
#if DEBUG_Switch
freopen("C:\code\input.txt", "r", stdin);
#endif
//freopen("C:\code\output.txt","r",stdin);
n = readint();
k = readint();
repd(i, 1, n)
{
repd(j, 1, k)
{
a[i][j] = readint();
}
}
build(1, 1, n);
int q = readint();
while (q--)
{
int op = readint();
if (op == 1)
{
int id = readint();
repd(j, 1, k)
{
a[id][j] = readint();
}
update(1, 1, n, id);
} else
{
int l = readint();
int r = readint();
for (int i = 0; i < (1 << k); ++i)
{
f[i] = -1e18;
}
query(1, 1, n, l, r);
ll ans = -1e18;
for (int i = 0; i < (1 << k); ++i)
{
int opponent =((1 << k)-1)^i;
ans = max(ans, f[i] + f[opponent]);
}
printf("%lld
", ans );
}
}
return 0;
}