[Educational Codeforces Round 56 (Rated for Div. 2)] —G. Multidimensional Queries（二进制状压，线段树）

[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\(\text 2\)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\(\sum \limits_{i = 1}^{k} |a_{x, i} - a_{y, i}|=max(\sum_{i=1}^{k}Q_i*a_i+\sum_{i=1}^{k}(!Q_i)*b_i,Q_i\in{\{-1,1\}})\)。

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\(\text 1\)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), '\0', 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("\n");}}}
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("\n");}}}
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 ? '\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 ? '\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\n",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\n",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\n", ans );
        }
    }
    return 0;
}