HDU 1114 PiggyBank

http://acm.hdu.edu.cn/showproblem.php?pid=1114

在0/1背包中有说，当只有f[0]=0,其他值为-∞时,可以求到恰好是装满时的最大值.

这题应用这个原理,即只有f[0]=0,其他的都不为0,应该讲它设置为+∞,则恰好可以求到装满时的最小指.

#include <iostream>
#define maxn 10005
#define M 505
using namespace std;
int ans[maxn], v[M], w[M];
int main()
{
    int t, n, m, i, j, a;
    cin >> t;
    while(t--)
    {
        cin >> a >> m;
        m -= a;
        cin >> n;
        for(i = 0; i < n; i++)
        cin >> w[i] >> v[i];
        for(i = 0; i <= m; i++)
          ans[i] = 0x7fffffff;
        ans[0] = 0;
        for(i = 0; i < n; i++)
        for(j = v[i]; j <= m; j++)
        {
            if( ans[j-v[i]]+w[i] < ans[j] && ans[j-v[i]] != 0x7fffffff)
               ans[j] = ans[j-v[i]]+w[i]; 
        }
        if(ans[m] == 0x7fffffff) cout << "This is impossible." << endl;
        else cout << "The minimum amount of money in the piggy-bank is " << ans[m] << "." << endl;
    }
    return 0;
}