Two Robots
Problem Statement :
You have a warehouse with M containers filled with an infinite number of candies. The containers are arranged in a single row, equally spaced to be 1 meter apart. You also have 2 robots that can pick up 1 piece of candy and transport it between any two containers. The robots take instructions in the form of queries consisting of two integers, Ma and Mb, respectively. To execute a query, a robot travels to container Ma, picks up 1 candy, transports it to container Mb, and then stops at Mb until it receives another query. Calculate the minimum total distance the robots must travel to execute N queries in order. Note: You choose which robot executes each query. Input Format The first line contains a single integer, T (the number of test cases); each of the T test cases is described over N+1 lines. The first line of a test case has two space-separated integers, M (the number of containers) and N (the number of queries). The N subsequent lines each contain two space-separated integers, Ma and Mb, respectively; each line Ni describes the ith query. Constraints 1 <= T <= 50 1 < M <= 1000 1 <= N <= 1000 1 <= a,b <= M Ma != Mb Output Format On a new line for each test case, print an integer denoting the minimum total distance that the robots must travel to execute the queries in order.
Solution :
Solution in C :
In C++ :
#include<bits/stdc++.h>
#include<iostream>
using namespace std;
#define fre freopen("0.in","r",stdin);freopen("0.out","w",stdout)
#define abs(x) ((x)>0?(x):-(x))
#define MOD 1000000007
#define fi first
#define se second
#define LL signed long long int
#define scan(x) scanf("%d",&x)
#define print(x) printf("%d\n",x)
#define scanll(x) scanf("%lld",&x)
#define printll(x) printf("%lld\n",x)
#define rep(i,from,to) for(int i=(from);i <= (to); ++i)
#define pii pair<int,int>
vector<int> G[2*100000+5];
int fdp[1005][1005];
LL dp[1005][1005], T=0;
pii Q[1000+5];
LL cost(int i,int j){
if(i==0)
return abs(Q[j].fi-Q[j].se);
return abs(Q[j].fi-Q[j].se) + abs(Q[j].fi-Q[i].se);
}
int N;
LL rec(int i,int j){
if(i==N or j==N)return 0;
if(fdp[i][j]==T)
return dp[i][j];
fdp[i][j] = T;
int k = max(i,j) + 1;
return dp[i][j] = min(rec(k,j)+cost(i,k),rec(i,k)+cost(j,k));
}
int main(){
//fre;
int t, M;
cin>>t;
while(t--){
T++;
cin>>M>>N;
for(int i=1;i<=N;++i){
scan(Q[i].fi);
scan(Q[i].se);
}
printll(rec(0,0));
}
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args){
int t = ni();
for(int i=0; i<t; i++){
solve();
}
}
static void solve(){
int m = ni(); int n = ni();
long[] mindis = new long[m+1];
long[] temp2;
long[] temp = new long[m+1];
Arrays.fill(mindis, Long.MAX_VALUE);
int curpos = 0;//current position
int ma, mb;
long d;
mindis[0] = 0;
for(int i=0; i<n; i++){
ma = ni(); mb = ni();
Arrays.fill(temp, Long.MAX_VALUE);
for(int j=0; j<=m; j++){
if(mindis[j] == Long.MAX_VALUE) continue;
d = mindis[j] + Math.abs(mb-ma);
if(j == 0)
temp[curpos] = Math.min( d , temp[curpos]);
else
temp[curpos] = Math.min( d + Math.abs(ma-j), temp[curpos]);
if(curpos == 0)
temp[j] = Math.min( d, temp[j]);
else
temp[j] = Math.min( d + Math.abs(ma-curpos) , temp[j]);
}
curpos = mb;
temp2 = mindis;
mindis = temp;
temp = temp2;
}
long min = mindis[0];
for(int i=0; i<=m; i++) min = Math.min(min, mindis[i]);
System.out.println(min);
}
static Scanner sc = new Scanner(System.in);
static int ni() { return sc.nextInt(); }
static void print(Object... objs){
System.out.println(Arrays.deepToString(objs));
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int t, t1;
scanf("%d", &t);
for (t1 = 0; t1 < t; t1++) {
int m, n, i, j, a, b;
scanf("%d %d", &m, &n);
int ar[m+1], r2 = 0, temp, min;
for (j = 0; j <= m; j++) ar[j] = 0;
for (i = 0; i < n; i++) {
scanf("%d %d", &a, &b);
min = ar[0] + abs(a - b);
for (j = 1; j <= m; j++) {
if (ar[j] == 0) continue;
temp = ar[j] + abs(j - a) + abs(a - b);
if (temp < min) min = temp;
}
if (r2 == 0) temp = abs(a - b);
else temp = abs(r2 - a) + abs(a - b);
ar[0] += temp;
for (j = 1; j <= m; j++) {
if (ar[j] == 0) continue;
ar[j] += temp;
}
if (ar[r2] == 0 || ar[r2] > min) ar[r2] = min;
r2 = b;
}
min = ar[0];
for (j = 1; j <= m; j++) if (ar[j] != 0 && ar[j] < min) min = ar[j];
printf("%d\n", min);
}
return 0;
}
In Python3 :
from collections import defaultdict
def dist(a, b):
if -1 in (a, b): return 0
return abs(a - b)
for _ in range(int(input())):
m, n = [int(x) for x in input().split()]
a, b = [int(x) for x in input().split()]
dp = {-1: abs(b-a)}
for _ in range(n-1):
newa, newb = [int(x) for x in input().split()]
d = abs(newa-newb) + dist(b, newa)
d2 = min(v + dist(k, newa) + abs(newa-newb) for k, v in dp.items())
for k in dp:
dp[k] += d
if b not in dp: dp[b] = float('inf')
dp[b] = min(dp[b], d2)
#print(dp)
b = newb
print(min(dp.values()))
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