[LeetCode] Minimum Total Distance Traveled

2463. Minimum Total Distance Traveled

There are some robots and factories on the X-axis. You are given an integer array robot where robot[i] is the position of the ith robot. You are also given a 2D integer array factory where factory[j] = [positionj, limitj] indicates that positionj is the position of the jth factory and that the jth factory can repair at most limitj robots.

The positions of each robot are unique. The positions of each factory are also unique. Note that a robot can be in the same position as a factory initially.

All the robots are initially broken; they keep moving in one direction. The direction could be the negative or the positive direction of the X-axis. When a robot reaches a factory that did not reach its limit, the factory repairs the robot, and it stops moving.

At any moment, you can set the initial direction of moving for some robot. Your target is to minimize the total distance traveled by all the robots.

Return the minimum total distance traveled by all the robots. The test cases are generated such that all the robots can be repaired.

Note that

  • All robots move at the same speed.
  • If two robots move in the same direction, they will never collide.
  • If two robots move in opposite directions and they meet at some point, they do not collide. They cross each other.
  • If a robot passes by a factory that reached its limits, it crosses it as if it does not exist.
  • If the robot moved from a position x to a position y, the distance it moved is |y - x|.
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#include <bits/stdc++.h>

#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")//Comment optimisations for interactive problems (use endl)
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

using namespace std;

struct PairHash {inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }};

// speed
#define Code ios_base::sync_with_stdio(false);
#define By ios::sync_with_stdio(0);
#define Sumfi cout.tie(NULL);

// alias
using ll = long long;
using ld = long double;
using ull = unsigned long long;

// constants
const ld PI = 3.14159265358979323846; /* pi */
const ll INF = 1e18;
const ld EPS = 1e-9;
const ll MAX_N = 202020;
const ll mod = 998244353;

// typedef
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef array<ll,3> all3;
typedef array<ll,5> all5;
typedef vector<all3> vall3;
typedef vector<all5> vall5;
typedef vector<ld> vld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef pair<string, string> pss;
typedef vector<pss> vpss;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef unordered_set<ll> usll;
typedef unordered_set<pll, PairHash> uspll;
typedef unordered_map<ll, ll> umll;
typedef unordered_map<pll, ll, PairHash> umpll;

// macros
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,m,n) for(ll i=n;i>=m;i--)
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define INF(a) memset(a,0x3f3f3f3f3f3f3f3fLL,sizeof(a))
#define ASCEND(a) iota(all(a),0)
#define sz(x) ll((x).size())
#define BIT(a,i) (a & (1ll<<i))
#define pyes cout<<"Yes\n";
#define pno cout<<"No\n";
#define pneg1 cout<<"-1\n";
#define ppossible cout<<"Possible\n";
#define pimpossible cout<<"Impossible\n";
#define TC(x) cout<<"Case #"<<x<<": ";
#define X first
#define Y second

// utility functions
template <typename T>
void print(T &&t) { cout << t << "\n"; }
template<typename T>
void printv(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<" ";cout<<"\n";}
void fileIO() {freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);}
void hackerCupIO() {freopen("/Users/summerflower/Downloads/worklife_balance_chapter_1_input (1).txt", "r", stdin); freopen("/Users/summerflower/Downloads/solution.txt","w",stdout);}
void readf() {freopen("", "rt", stdin);}
template<typename T>
void readv(vector<T>& v){rep(i,0,sz(v)) cin>>v[i];}
template<typename T, typename U>
void readp(pair<T,U>& A) {cin>>A.first>>A.second;}
template<typename T, typename U>
void readvp(vector<pair<T,U>>& A) {rep(i,0,sz(A)) readp(A[i]); }
void readvall3(vall3& A) {rep(i,0,sz(A)) cin>>A[i][0]>>A[i][1]>>A[i][2];}
void readvvll(vvll& A) {rep(i,0,sz(A)) readv(A[i]);}

struct Combination {
vll fac, inv;
ll n, MOD;

ll modpow(ll n, ll x, ll MOD = mod) { if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

Combination(ll _n, ll MOD = mod): n(_n + 1), MOD(MOD) {
inv = fac = vll(n,1);
rep(i,1,n) fac[i] = fac[i-1] * i % MOD;
inv[n - 1] = modpow(fac[n - 1], MOD - 2, MOD);
rrep(i,1,n - 2) inv[i] = inv[i + 1] * (i + 1) % MOD;
}

ll fact(ll n) {return fac[n];}
ll nCr(ll n, ll r) {
if(n < r or n < 0 or r < 0) return 0;
return fac[n] * inv[r] % MOD * inv[n-r] % MOD;
}
};

// geometry data structures
template <typename T>
struct Point {
T y,x;
Point(T y, T x) : y(y), x(x) {}
Point() {}
void input() {cin>>y>>x;}
friend ostream& operator<<(ostream& os, const Point<T>& p) { os<<p.y<<' '<<p.x<<'\n'; return os;}
Point<T> operator+(Point<T>& p) {return Point<T>(y + p.y, x + p.x);}
Point<T> operator-(Point<T>& p) {return Point<T>(y - p.y, x - p.x);}
Point<T> operator*(ll n) {return Point<T>(y*n,x*n); }
Point<T> operator/(ll n) {return Point<T>(y/n,x/n); }

Point<T> rotate(Point<T> center, ld angle) {
ld si = sin(angle * PI / 180.), co = cos(angle * PI / 180.);
ld y = this->y - center.y;
ld x = this->x - center.x;

return Point<T>(y * co - x * si + center.y, y * si + x * co + center.x);
}
};

template<typename T>
struct Circle {
Point<T> center;
T radius;
Circle(T y, T x, T radius) : center(Point<T>(y,x)), radius(radius) {}
Circle() {}

void input() {
center = Point<T>();
center.input();
cin>>radius;
}

bool circumference(Point<T> p) {
return (center.x - p.x) * (center.x - p.x) + (center.y - p.y) * (center.y - p.y) == radius * radius;
}

bool intersect(Circle<T> c) {
T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
return (radius - c.radius) * (radius - c.radius) <= d and d <= (radius + c.radius) * (radius + c.radius);
}

bool include(Circle<T> c) {
T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
return d <= radius * radius;
}
};

ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
all3 __exgcd(ll x, ll y) { if(!y) return {x,1,0}; auto [g,x1,y1] = __exgcd(y, x % y); return {g, y1, x1 - (x/y) * y1}; }
ll __lcm(ll x, ll y) { return x / __gcd(x,y) * y; }
ll modpow(ll n, ll x, ll MOD = mod) { n%=MOD; if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

ll source = 202, sink = 203;
ll path[222], cost[222][222], flo[222][222], cap[222][222];
vll adj[222];

class Solution {
void connect(ll u, ll v, ll capacity, ll weight) {
adj[u].push_back(v);
adj[v].push_back(u);

cap[u][v] += capacity;

cost[u][v] += weight;
cost[v][u] += -weight;
}
bool bfs(ll u, ll v) {
vll w(222, INF);
bool inc[222];
memset(path, 0, sizeof path);
memset(inc, 0, sizeof inc);
queue<ll> q;
q.push(u);
w[u] = 0;

while(!q.empty()) {
auto n = q.front(); q.pop();
inc[n] = false;
for(auto& m : adj[n]) {
if(cap[n][m] - flo[n][m] > 0 and w[n] + cost[n][m] < w[m]) {
w[m] = w[n] + cost[n][m];
path[m] = n;
if(!inc[m]) {
inc[m] = true;
q.push(m);
}
}
}
}

return path[v] != 0;
}
public:
long long minimumTotalDistance(vector<int>& A, vector<vector<int>>& B) {
ZERO(cost); ZERO(flo); ZERO(cap);
rep(i,0,222) adj[i].clear();
ll gap = 100;
rep(i,0,sz(A)) {
rep(j,0,sz(B)) {
connect(i, j + gap, 1, abs(A[i] - B[j][0]));
}
}
rep(i,0,sz(A)) {
connect(source, i, 1, 0);
}
rep(i,0,sz(B)) {
connect(gap + i, sink, B[i][1], 0);
}
ll res = 0, flow = 0;
while(bfs(source, sink)) {
ll v = sink, mi = INF;
while(source != v) {
ll u = path[v];
mi = min(mi, cap[u][v] - flo[u][v]);
v = u;
}
v = sink;
flow += mi;
while(source != v) {
ll u = path[v];
res += mi * cost[u][v];
flo[u][v] += mi;
flo[v][u] -= mi;
v = u;
}
}
return res;
}
};
Author: Song Hayoung
Link: https://songhayoung.github.io/2022/11/06/PS/LeetCode/minimum-total-distance-traveled/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.