Bolivia

ACM ICPC Latin America 2011 – A – Army Buddies

This problem is a simple case of elimination for ranges, given a range, we need remove it and obtain it’s neighbors. the best option for this case is a structure of linked list.

For example the second case in the problem:
S=10 B=4   ===> * 1 2 3 4 5 6 7 8 9 10 *
2 5        ===> * 1(2 3 4 5)6 7 8 9 10 *
                * 1 6 7 8 9 10 *
6 9        ===> * 1        (6 7 8 9)10 *
                * 1 10 *
1 1        ===> *(1)                10 *
                * 10 *
10 10      ===> *                  (10)*
                * *

This is my implementation and you can see in live archive’s statistic (Run Time:0.104).

/* 
 * @author: vudduu - Edwin Guzman
 * @problem: Problem A - Army Buddies
 * @contest: ACM ICPC Latin America Regional 2011
 */
#include <iostream>
#include <sstream>
#include <utility>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <functional>
#include <algorithm>
#include <numeric>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
using namespace std;

#define FOR(i,a,b)  for(int i=(a),_##i=(b);i<_##i;++i)
#define F(i,a)      FOR(i,0,a)
#define ALL(x)      x.begin(),x.end()
#define PB          push_back
#define MP          make_pair
#define S           size()
typedef long long   LL;

int vleft[100001], vright[100001];

int main(){
   int n, x, y, m;
   scanf("%d %d", &n, &m);
   while(n){
      FOR(i, 1, n+1){
         vleft[i] = i-1;
         vright[i] = i+1;
      }
      F(i, m){
         scanf("%d %d", &x, &y);
         if(vleft[x] == 0) printf("* ");
         else printf("%d ", vleft[x]);
         if(vright[y] == n+1) printf("*n");
         else printf("%dn", vright[y]);
         vleft[ vright[y]] = vleft[x];
         vright[ vleft[x]] = vright[y];
      }
      printf("-n");
      scanf("%d %d", &n, &m);
   }
}

ACM ICPC Latin America 2011 – I – In Braille

/* 
 * @author: vudduu - Edwin Guzman
 * @problem: Problem I - In Braille
 * @contest: ACM ICPC Latin America Regional 2011
 */
#include <iostream>
#include <sstream>
#include <utility>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <functional>
#include <algorithm>
#include <numeric>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
using namespace std;

#define FOR(i,a,b)  for(int i=(a),_##i=(b);i<_##i;++i)
#define F(i,a)      FOR(i,0,a)
#define ALL(x)      x.begin(),x.end()
#define PB          push_back
#define MP          make_pair
#define S           size()
typedef long long   LL;

string s1[] = {".*", "*.", "*.", "**", "**", "*.", "**", "**", "*.", ".*"};
string s2[] = {"**", "..", "*.", "..", ".*", ".*", "*.", "**", "**", "*."};

char ch[305];
char num[105];
char s[3][305];

int it;

void lee(int & n){
   it = n = 0;
   gets(ch);
   while(ch[it]) n = n*10 + ch[it++]-'0';
}

int main(){
   int n;
   lee(n);
   while(n){
      gets(ch);
      it = 0;
      if(ch[0] == 'S'){
         gets(num);
         for(int i=0; i<n ;i++){
            if(i) s[0][it] = ' ', s[1][it] = ' ', s[2][it++] = ' ';
            char numaux = num[i]-'0';
            s[0][it] = s1[ numaux ][0];
            s[1][it] = s2[ numaux ][0];
            s[2][it++] = '.';
            s[0][it] = s1[ numaux ][1];
            s[1][it] = s2[ numaux ][1];
            s[2][it++] = '.';
         }
         s[0][it] = s[1][it] = s[2][it] = 0;
         puts(s[0]);
         puts(s[1]);
         puts(s[2]);
      }
      else{
         for(int i=0; i<3 ;i++)
            gets(s[i]);
         for(int i=0; i<n ;i++){
            int it1 = i*3, j;
            for(j=0; j<10 ;j++){
               if(s[0][it1] == s1[j][0]) if(s[0][it1+1] == s1[j][1])
                  if(s[1][it1] == s2[j][0]) if(s[1][it1+1] == s2[j][1])
                     break;
            }
            num[it++] = j+'0';
         }
         num[it] = 0;
         puts(num);
      }
      lee(n);
   }
}

ACM ICPC Latin America 2011 – K – King’s Poker

/* 
 * @author: vudduu - Edwin Guzman
 * @problem: Problem K - King's Poker
 * @contest: ACM ICPC Latin America Regional 2011
 */
#include <iostream>
#include <sstream>
#include <utility>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <functional>
#include <algorithm>
#include <numeric>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
using namespace std;

#define FOR(i,a,b)  for(int i=(a),_##i=(b);i<_##i;++i)
#define F(i,a)      FOR(i,0,a)
#define ALL(x)      x.begin(),x.end()
#define PB          push_back
#define MP          make_pair
#define S           size()
typedef long long   LL;

int it;
char ch[30];

int main(){
  int v[3];
  scanf("%d %d %d", &v[0], &v[1], &v[2]);
  while(v[0]){
    sort(v, v+3);
    if(v[0] == v[1] && v[0] == v[2]){
      if(v[0] == 13) printf("*n");
      else printf("%d %d %dn", v[0]+1, v[0]+1, v[0]+1);
    }
    else if(v[0] == v[1]){
      if(v[2] == 13) printf("1 %d %dn", v[0]+1, v[0]+1);
      else printf("%d %d %dn", v[0], v[0], v[2]+1);
    }
    else if(v[1] == v[2]){
      if(v[0] == v[1]-1){
        if(v[1] == 13) printf("1 1 1n");
        else printf("%d %d %dn", v[2], v[2], v[0]+2);
      }
      else printf("%d %d %dn", v[0]+1, v[2], v[2]);
    }
    else printf("1 1 2n");
    scanf("%d %d %d", &v[0], &v[1], &v[2]);
  }
}