It’s time to blog again.. Now i will post some small-writeup about the problems were used at this contest..

Well.., about today small-writeup, i will not post my source code anymore. Because
in practice we are sure that reading the source code at first sight will make us
bored and no creativity to implement our own solution..

Here is me at standing page :

Okay then, i will only put my ideas in this writeup. So, enjoy it..

Problem A : Letter (Accepted 1st try)
It is a very nice problem at the beginning. We are given rectangular board, we should take some rectangular board (possibly smaller) that cover all the shaded square. My idea is to make 4 lines, and take their intersection to be the left topmost and the right bottom, so we can make a rectangle with that. After that, print the solution. It can be done in O(n * m) time. Since the n and m is small, the algorithm above will be run in-time.

Problem B : Young Photographer (Accepted 1st try)
This problem is about “array differencing” problem. But since the constraints were small enough, i decided to code it in the worst complexity (maybe). My idea is just make an array of positions, then increment it one-by-one if there are sports-man at that position. The complexity of this solution is O(n ^ 2) i think. The solution of this problem is to compare each position that contain n sports-man at there and the initial position of the photographer, take the minimum.

Problem C : Four Segments (Accepted 3rd try)
This is a simple geometry problem. What did i do is to generate permutation of the segments, and decide if the rectangle is exist there. The complexity is O(4!), since there are 4 segments given. I did Wrong Answer 2 times because a silly mistakes. And it is time-consuming problem.

Problem D : Two Paths (Wrong Answer)
I can’t solve this problem yet. But actually my idea is to select some vertex and make it a “divider”. But i think that’s the wrong idea, so never mind.. Maybe anyone can help me to solve this? I am very sure that i have no idea how to solve it because of there are some algorithms that i never learn before.

Problem E : Camels (Accepted 1st try)
Well, this is my favorite problem in this contest. The combination about geometry stuffs and dynamic programming. I used 5-dimensions table of DP to solve it. Maybe anyone have a better complexity or memory? I spent many times to write the code, because the implementation was quite time-consuming.

Yeah, that’s all..
Feedbacks are welcome then.. Please comment if you like this post or
want to give some tips, etc?

Best Regards,
Wilbert Liu

But now i have lots of free time, because i’ve already finished such “busiest” things at my university time. And i think, it’s time to fill this blog with some new stories, sketch, and something funny that will entertain your day (i hope)

The first thing that i should consider about posting some “algorithm write-up” is i will not post the code anymore. The reason behind that is obvious, i want the reader to think about the implementation by themselves. Many times i found that reading other implementation at first made us not creative anymore.

Okay then, let’s wait for the new post..

So i can’t update this blog regularly. But now i have a time to rest, since my research paper was submitted, so that i just wait for the thesis defense schedule around May..

Well.., i spent my time with my lovely girlfriend, and doing something funny (of course) at SPOJ (you know that well). Yeah, i solved problems at there..

I think i will share something tomorrow.., something about problems of course..

Stay tuned!

Standings

Problem A :

It is a nice problem (i think). Given W, the weight of watermelon bought by the boys, we have to determine whether the watermelon can be divided into 2 parts with each of them having even number of kilos, or not. At first, brute force came up in my mind, but then i found a really nice solution for this problem. The first observation is, assume the watermelon with weight w can be divided into 2 parts, so, w = 2x + 2y = 2(x + y), because each part must have even value. From the first observation, we can see that if w can be divided into 2 parts, then w must be even value. Then, it’s easy to see that 2x + 2y must be >= 4, because the lowest positive even number is 2, so 2 + 2 = 4. Here is my implementation :

#include <cstdio>

using namespace std;

int n;

int main()
{
    while(scanf("%d", &n) != EOF)
        printf(n % 2 == 0 && n > 2 ? "YES\n" : "NO\n");
    return 0;
}

Problem B :
There are several approaches to solve this problem, but i used to solve this problem with Dynamic Programming, since i can’t proof other approaches like Greedy, etc. Post on comment if you have any questions about my DP.. . So, without doubt, here is my implementation for this problem :

#include <algorithm>
#include <cstdio>
#include <cstring>

using namespace std;

int d, sumTime, path[30];
bool dp[30][241], vis[30][241];
pair<int, int> tme[30];

bool doit(int pos, int t)
{
    if(pos == d)
    {
        if(t == 0)
            return true;
        else
            return false;
    }
    bool &ret = dp[pos][t];
    if(!vis[pos][t])
    {
        vis[pos][t] = true;
        ret = false;
        for(int i = tme[pos].first; i <= tme[pos].second; ++i)
        {
            bool check = false;
            if(t >= i)
                check = doit(pos + 1, t - i);
            if(check)
                path[pos] = i;
            ret |= check;
        }
    }
    return ret;
}

int main()
{
    while(scanf("%d %d", &d, &sumTime) != EOF)
    {
        memset(vis, false, sizeof(vis));
        for(int i = 0; i < d; ++i)
            scanf("%d %d", &tme[i].first, &tme[i].second);
        for(int i = tme[0].first; i <= tme[0].second; ++i)
        {
            bool check = false;
            if(sumTime >= i)
                check = doit(1, sumTime - i);
            if(check)
                path[0] = i;
        }
        if(path[0] >= tme[0].first && path[0] <= tme[0].second)
        {
            printf("YES\n%d", path[0]);
            for(int i = 1; i < d; ++i)
                printf(" %d", path[i]);
        }
        else
            printf("NO");
        printf("\n");
    }
    return 0;
}

Problem C :

From the first time i read the problem statement, i know it can be solved with map. I implemented it quickly and got AC at first time submit. Here is my implementation :

#include <cstdio>
#include <iostream>
#include <map>
#include <string>

using namespace std;

int n;
string s;
map<string, int> mp;

int main()
{
    while(scanf("%d", &n) != EOF)
    {
        mp.clear();
        getline(cin, s);
        for(int i = 0; i < n; ++i)
        {
            getline(cin, s);
            if(!mp[s])
            {
                ++mp[s];
                printf("OK\n");
            }
            else
            {
                printf("%s%d\n", s.c_str(), mp[s]);
                ++mp[s];
            }
        }
    }
    return 0;
}

Problem D :

I think this is the hardest problem compared to the other problems. At this problem i have to find the answer for maximum chain and i have to print the “path” of the answer. I just thought a simple way, because this problem is similar to Longest Increasing Subsequence problem. Printing path was quite easy job here. I got WA 2 times because i tried to proof Greedy solution for this problem. At first i was sure that my Greedy was true, but after submitting the solution, i got WA and i can’t found the couter example to disprove my solution. So, i turned out to think about DP to solve this problem. Here is my implementation :

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

struct env
{
    int w, h, id;

    bool operator < (const env &e) const
    {
        if(w < e.w || w == e.w && h < e.h)
            return true;
        return false;
    }
};

int n, w, h, dp[5000], path[5000];
env e[5000];

int main()
{
    while(scanf("%d %d %d", &n, &w, &h) != EOF)
    {
        int cnt = 0;
        for(int i = 0; i < n; ++i)
        {
            env tem;
            scanf("%d %d", &tem.w, &tem.h);
            if(tem.w > w && tem.h > h)
            {
                e[cnt].w = tem.w;
                e[cnt].h = tem.h;
                e[cnt].id = i + 1;
                ++cnt;
            }
        }
        sort(e, e + cnt);
        if(cnt > 0)
        {
            for(int i = 0; i < cnt; ++i)
            {
                dp[i] = 1;
                path[i] = i;
                for(int j = 0; j < i; ++j)
                    if(e[j].w < e[i].w && e[j].h < e[i].h && dp[i] < dp[j] + 1)
                    {
                        dp[i] = dp[j] + 1;
                        path[i] = j;
                    }
            }
            int res = 1, tip = 0;
            for(int i = 0; i < cnt; ++i)
            if(res < dp[i])
            {
                res = dp[i];
                tip = i;
            }
            vector<int> resPath;
            resPath.push_back(e[tip].id);
            while(path[tip] != tip)
            {
                resPath.push_back(e[path[tip]].id);
                tip = path[tip];
            }
            printf("%d\n%d", res, resPath[resPath.size() - 1]);
            for(int i = resPath.size() - 2; i >= 0; --i)
                printf(" %d", resPath[i]);
        }
        else
            printf("0");
        printf("\n");
    }
    return 0;
}

Feel free to ask anything here..
Best Regards,
Wilbert Liu

There are 5 problems with various difficulty, and i found myself solved the first two problems easily, and then i scratched my mind to solve the third problem, but i can’t found any solutions at that time.., but still thinking to find the right way to solve that one..

I will post my solution on the other problems if i can solve them later..

Here is my solutions to problem A and B

//A.cpp
#include <cstdio>
#include <iostream>
#include <string>

using namespace std;

string s;
int fr = 0, res = 0;

int main()
{
    while(getline(cin, s))
    {
        if(s[0] == '+')
            ++fr;
        else if(s[0] == '-')
            --fr;
        else
        {
            for(int i = 0; i < s.size(); ++i)
                if(s[i] == ':')
                {
                    if(i + 1 < s.size())
                        res += (s.size() - (i + 1)) * fr;
                    break;
                }
         }
     }
     printf("%d\n", res);
     return 0;
}

//B.cpp
#include <algorithm>
#include <cstdio>
#include <iostream>
#include <string>

using namespace std;

string s, w[1000];
int longest = 0, turn = 0;

int main()
{
    int cnt = 0;
    while(getline(cin, s))
    {
        longest = max(longest, (int) s.size());
        w[cnt] = s;
        ++cnt;
    }
    for(int i = 0; i < longest + 2; ++i)
        printf("*");
    printf("\n");
    for(int i = 0; i < cnt; ++i)
    {
        printf("*");
        int leng = longest - w[i].size();
        if(leng % 2 == 0)
        {
            for(int j = 0; j < leng / 2; ++j)
            printf(" ");
            printf("%s", w[i].c_str());
            for(int j = 0; j < leng / 2; ++j)
                printf(" ");
        }
        else
        {
            for(int j = 0; j < leng / 2 + turn; ++j)
                printf(" ");
            printf("%s", w[i].c_str());
            for(int j = 0; j < leng / 2 + !turn; ++j)
                printf(" ");
            turn = !turn;
        }
        printf("*\n");
    }
    for(int i = 0; i < longest + 2; ++i)
        printf("*");
    printf("\n");
    return 0;
}

After few moments, i decide to do blogging again.. I think it’s important to share all of my life to the other people.. My daily life, relationship, vision, target, etc.. How about my previous blog? Hmm.., maybe i will start to ignore that, since this blog will contain tech things like programming, math, algorithm, etc..

Well, i try to post all things here with English, because i am learning English right now.. :p So, forgive me if there are typos here.. I’d be glad if someone remind me about my mistakes..

Okay, enjoy my blog then, i will try to update it regularly if i have enough time..

Best Regards,

Wilbert Liu