Or at least add another version of the problem with huge labyrinths.
I had a lot of fun with this problem, and was surprised that a solution which is far from perfect can work instantly with input 200x200.

3
.##
###
##.

In C#, I was getting TLE using LinkedList, so I rewrote everything using an array of 2*10^6 elements. It passed in 1.7s (thank god), but that's still very slow. How do people get times like 0.2s?

Don't be trapped by the title of this problem...it would just make it more complicated

Edited by author 06.02.2016 13:30

Edited by author 06.02.2016 13:38

int num,n,i,sum=0;
     // read n;

//case 1
       if(n>0){
            for(i=1;i<=n;i++){
            sum=sum+i;
        }
         printf("%d",sum);
        }
//case 2
        if(n<0){
             for(i=(-1);i>=n;i--){

            sum=sum+i;
        }
            sum=sum+1;
            printf("%d",sum);
        }
//case 3
        else if(n==0){
            printf("1");
        }

        }

Using the same code, and only switching the order of the 2 input strings ...
    scanf("%s", strr);
    scanf("%s", strl);
leads to WA5

switching the order (shouldn`t make any difference)
    scanf("%s", strl);
    scanf("%s", strr);
and I get WA6

I tried everything suggested in the discussion.
Each test works, but when I submit, I get WA2.

Can you please add more tests?

Looks like stations can overlap (have same coordinates) while not being connected with underground!

What is my timus handle ? in a place I give my username and they decline that and say that timus handle should be a number. What is my number? how can I find the ID?



Edited by author 14.12.2022 22:58

cost, fun1, fun2 = map(int,input().split())
fri, flat2 = [], []
for i in range(int(input())):
        fri.append(list(map(int,input().split())))

temp1, temp2 = [-1,-1], [-1, -1, -1] #веселье, квартрира, сосед
for i in range(int(input())) :
        a, b, c = map(int,input().split())
        if a == 2 :
                flat2.append([b,c,i+1]) # стоимость, веселье, номер
        if fun1 + c > temp1[0] and cost >= b :
                temp1 = [fun1+c, i+1]

for i in range(len(flat2)) :
        b, c, k  = flat2[i][0], flat2[i][1], flat2[i][2]
        for j in range(len(fri)) :
                if fri[j][0] >= b/2 and cost >= b/2 and fun2+fri[j][1]+c > temp2[0] :
                        temp2 = [fun2+fri[j][1]+c, j+1, k]

if temp1[0] == -1 and temp2[0] == -1 :
        print('Forget about apartments. Live in the dormitory.')
elif temp1[0] > temp2[0] :
        print('You should rent the apartment #'+str(temp1[1])+ ' alone.')
else :
        print('You should rent the apartment #'+str(temp2[2])+' with the friend #'+str(temp2[1])+'.')

Edited by author 16.09.2016 21:53

hi, this is O(1) complexity problem, you might want to decrease the Difficulty i guess

to others: if you have found an extremum skip next point

AC in 0.015 is possible without any precalculations.
Make a function that returns the quiet days until day N

The answer is F(r) - F(l).
The is a special case if l has to be counted - that happens if F(l) != F(l-1) (in other words F(l) is part of the sequence we are looking for)

If l=1, just return F(r)

The implementation of F(n) can be simple if you find the right approach.
It's simple to count how many days remain in the sequence after eliminating each 2nd, then each 3rd, and so on.

Out of all shortest path I am finding min of maximum distance I have to travel for every level.





#include<bits/stdc++.h>
using namespace std;

void bfs(int u,vector<int> &d,const vector<vector<int>> adj){
    d[u] = 0;
    queue<int> q;
    q.push(u);

    while(!q.empty()){
        int u = q.front();
        q.pop();

        for(auto v:adj[u]){
            if(d[v]>d[u]+1){
                d[v]=d[u]+1;
                q.push(v);
            }
        }
    }
}

int main(){
    int n,m;
    cin>>n>>m;
    vector<vector<int>> adj(n);

    for(int i=0;i<m;i++){
        int u,v;
        cin>>u>>v;
        adj[u-1].push_back(v-1);
        adj[v-1].push_back(u-1);
    }

    vector<int> ds(n,1e9),df(n,1e9),dr(n,1e9);
    int s,f,r;
    cin>>s>>f>>r;
    s--,f--,r--;
    bfs(s,ds,adj);
    bfs(f,df,adj);
    bfs(r,dr,adj);

    vector<vector<int>> l(ds[f]+1);

    for(int i=0;i<n;i++){
        if(ds[i]+df[i]==ds[f]){
            l[ds[i]].push_back(i);
        }
    }
    int ans = INT_MAX;
    for(int i=0;i<=ds[f];i++){
        int a = INT_MIN;
        for(auto x: l[i]){
            a = max(a,dr[x]);
        }
        ans = min(a,ans);
    }

    cout<<ans;

    return 0;
}

Just think of an increasing sequence. The program caculates n+1 times for an increasing sequence with n elements. And then divide it into two subsequences. One is the leftmost element, and anthoer is the left elements. It ends at the length of sequence is 2. Thus, the sum is (n+1) + n + (n-1) + ... + 3 = (n+4)(n-1)/2 = (n*n + 3*n - 4)/2. So output 1 to n is ok.

union message
{
    char bytes[4];
    unsigned value;
    void reverse()
    {
        swap(bytes[0], bytes[3]);
        swap(bytes[1], bytes[2]);
    }
};

#include <iostream>

using namespace std;

int main()
{
    int a;
    cin>>a;
    if (a>=7)
        cout<<"Yes";
    if (a<7)
        cout<<"No";
    return 0;
}

Consider the case in which n < i, j
That helped me to get AC

Is the first test from conditions? If answer is "yes", then why the system say "WA1"? The program works correctly on my computer. Maybe, I misunderstood something important, and my output has wrong format? Help me, please.


Edited by author 05.01.2013 15:42