HDOJ 1114 Piggy-Bank 动态规划

题目
反向多重背包 ，当背包恰好装满时，求最小价值

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

typedef long long ll;

int T, E, F, N;
int val[510], wei[510], dp[10010];

int main(void)
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> T;
    while (T--) {
        cin >> E >> F >> N;
        int V = F - E;
        for (int i = 1; i <= N; i++) { 
            cin >> val[i] >> wei[i];
        }
        memset(dp, 127, sizeof(dp));
        dp[0] = 0;
        for (int i = 1; i <= N; i++) {
            for (int j = wei[i]; j <= V; j++) {
                dp[j] = min(dp[j], dp[j - wei[i]] + val[i]);
            }
        }
        if (dp[V] <= 25000000) {
            cout << "The minimum amount of money in the piggy-bank is " << dp[V] << ".\n";
        }
        else {
            cout << "This is impossible.\n";
        }

    }
    return 0;
}