UVa 10306 - e-Coins

　　题目大意：现在有一种新型货币，它的价值分为传统价值x和IT价值y，价值计算方式为sqrt(x*x+y*y)，现给一些类型的货币和要达到的目标价值，计算达到目标所需的最少货币数目。注意计算方法是sqrt(所有x和y和的平方和)。

　　二维完全背包问题？

 #include <cstdio>
 #include <climits>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 #define MAXN 310

 int dp[MAXN][MAXN];
 int x[], y[];

 int main()
 {
 #ifdef LOCAL
     freopen("in", "r", stdin);
 #endif
     int T;
     scanf("%d", &T);
     while (T--)
     {
         int n, target;
         scanf("%d%d", &n, &target);
         for (int i = ; i < n; i++)
             scanf("%d%d", &x[i], &y[i]);
         for (int i = ;  i <= target; i++)
             for (int j = ; j <= target; j++)
                 dp[i][j] = INT_MAX;
         dp[][] = ;
         for (int i = ; i <= target; i++)
             for (int j = ; j <= target; j++)
                 for (int k = ; k < n; k++)
                     if (i-x[k] >=  && j-y[k]>=  && dp[i-x[k]][j-y[k]] != INT_MAX)
                         dp[i][j] = min(dp[i][j], dp[i-x[k]][j-y[k]]+);
         int ans = INT_MAX;
         int t = target * target;
         for (int i = ; i <= target; i++)
             for (int j = ; j <= target; j++)
                 if (i*i + j*j == t && dp[i][j] != INT_MAX)
                     ans = min(ans, dp[i][j]);
         if (ans != INT_MAX)  printf("%d\n", ans);
         else  printf("not possible\n");
     }
     return ;
 }