POJ 2082 The Fewest Coins

1.题目描述：点击打开链接

2.解题思路：本题是完全背包+多重背包的题目。根据题意，可以用dp[i]表示面值为i时候最少用到多少个硬币，根据题意不难发现，John给出硬币的情况是多重背包，而shopkeeper找零的情况是完全背包。这样，可以事先求出来相应的结果，最后扫描一遍求解最小值即可。不过这里有一个难点是如何确定枚举的上界。这里我根据感觉设置的是W+12000作为最大上界。

3.代码：

#include<iostream>
#include<algorithm>
#include<cassert>
#include<string>
#include<sstream>
#include<set>
#include<bitset>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<cctype>
#include<list>
#include<complex>
#include<functional>
using namespace std;

#define me(s) memset(s,0,sizeof(s))
typedef long long ll;
typedef pair <int,int> P;

const int N = 20000 + 100;
const int INF = 1e9;
int dp1
, dp2
;
int w
, num
;
int n, W;

void multi_package(int W, int*dp, int*C)
{
fill(dp, dp + W, INF);
dp[0] = 0;
for (int i = 0; i<n; i++)
{
int nu = C[i];
for (int k = 1; nu>0; k <<= 1)
{
int mul = min(nu, k);
for (int j = W; j >= w[i] * mul; j--)
dp[j] = min(dp[j], dp[j - w[i] * mul] + mul);
nu -= mul;
}
}
}

void complete_package(int W, int*dp)
{
fill(dp, dp + W, INF);
dp[0] = 0;
for (int i = 0; i<n; i++)
for (int j = w[i]; j<W; j++)
dp[j] = min(dp[j], dp[j - w[i]] + 1);
}

int main()
{
while (~scanf("%d%d", &n, &W))
{
for (int i = 0; i<n; i++)
scanf("%d", &w[i]);
for (int i = 0; i<n; i++)
scanf("%d", &num[i]);
multi_package(W + 12000, dp1, num);
complete_package(W + 12000, dp2);
int ans = INF;
for (int i = W; i<W + 12000; i++)
ans = min(ans, dp1[i] + dp2[i - W]);
if(ans==INF)puts("-1");
else printf("%d\n", ans);
}
}