The Fewest Coins POJ - 3260

題目傳送門

題意：你去一個商店購買一個價值爲T的物品，市場上有n種流通的硬幣，面值分別爲V1,V2..Vn (1<=Vi<=120)。John有Ci個面值爲Vi的硬幣(0<=Ci<=10000)。店主有無限多的硬幣，並總按最優方案找零。
問你最少使用多少個硬幣（包括店主找零的）

思路：完全揹包加多重揹包。

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

#define MAXN 200010
#define MAXE 210
#define INF 1000000000
#define MOD 10001
#define LL long long
#define ULL unsigned long long
#define pi 3.14159

using namespace std;

int dp[MAXN];
int v[MAXE];
int w[MAXE];
int vis[MAXN];


int main() {
    std::ios::sync_with_stdio(false);
    int n, m;
    cin >> n >> m;
    int max_v = 0;
    for (int i = 0; i < n; ++i) {
        cin >> v[i];
        max_v = max(max_v, v[i]);
        v[i + n] = -v[i];
    }
    for (int i = 0; i < n; ++i) {
        cin >> w[i];
    }
    memset(dp, -1, sizeof(dp));
    dp[0] = 0;
    for (int i = 0; i < n; ++i) {
        memset(vis, 0, sizeof(vis));
        for (int j = v[i]; j <= m + 30000; ++j) {
            if (dp[j - v[i]] != -1 && (dp[j - v[i]] + 1 < dp[j] || dp[j] == -1) && vis[j - v[i]] < w[i]) {
                dp[j] = dp[j - v[i]] + 1;
                vis[j] = dp[j - v[i]] + 1;
            }
        }
    }
    for (int i = n; i < 2 * n; ++i) {
        for (int j = m + 30000 + v[i]; j >= 0; --j) {
            if (dp[j - v[i]] != -1 && (dp[j - v[i]] + 1 < dp[j] || dp[j] == -1)) {
                dp[j] = dp[j - v[i]] + 1;
            }
        }
    }
    cout << dp[m] << endl;
    return 0;
}