12324 - Philip J. Fry Problem/12324.cpp

   1 // Equipo Poncho, Carriel y Ruana
   2 using namespace std;
   3 #include <algorithm>
   4 #include <iostream>
   5 #include <iterator>
   6 #include <sstream>
   7 #include <fstream>
   8 #include <cassert>
   9 #include <climits>
  10 #include <cstdlib>
  11 #include <cstring>
  12 #include <string>
  13 #include <cstdio>
  14 #include <vector>
  15 #include <cmath>
  16 #include <queue>
  17 #include <stack>
  18 #include <list>
  19 #include <map>
  20 #include <set>
  21
  22 template <class T> string toStr(const T &x)
  23 { stringstream s; s << x; return s.str(); }
  24 template <class T> int toInt(const T &x)
  25 { stringstream s; s << x; int r; s >> r; return r; }
  26
  27 #define For(i, a, b) for (int i=(a); i<(b); ++i)
  28 #define foreach(x, v) for (typeof (v).begin() x = (v).begin(); x != (v).end(); ++x)
  29 #define D(x) cout << #x " = " << (x) << endl;
  30
  31 const double EPS = 1e-9;
  32
  33 int cmp(double x, double y = 0, double tol = EPS) {
  34     return (x <= y + tol) ? (x + tol < y) ? -1 : 0 : 1;
  35 }
  36
  37 const int MAXN = 110;
  38 const int MAXB = MAXN * MAXN;
  39 int dp[MAXN][MAXB];
  40
  41 int time_spent[MAXN];
  42 int balls[MAXN];
  43
  44 int main(){
  45     int n;
  46     while (cin >> n) {
  47         if (n == 0) break;
  48         int totalBalls = 0;
  49         For(i, 0, n) {
  50             cin >> time_spent[i] >> balls[i];
  51             totalBalls += balls[i];
  52         }
  53         for (int b = 0; b <= totalBalls; ++b) dp[n][b] = 0;
  54         for (int i = n - 1; i >= 0; --i) {
  55             for (int b = 0; b <= totalBalls; ++b) {
  56                 dp[i][b] = dp[i + 1][b + balls[i]] + time_spent[i];
  57                 if (b > 0) {
  58                     int option = dp[i + 1][b - 1 + balls[i]] + time_spent[i] / 2;
  59                     if (option < dp[i][b]){
  60                         dp[i][b] = option;
  61                     }
  62                 }
  63             }
  64         }
  65         cout << dp[0][0] << endl;
  66     }
  67     return 0;
  68
  69 }