91. 存钱罐

题目链接

点此跳转

思路

完全背包问题。

令 $f[i][j]$ 为从前 $i$ 个物品中选且总体积等于 $j$ 的最小价值。

转移方程：$f[i][j]=min(f[i-1][j],f[i-1][j-v[i]]+w[i])$

边界条件：

$f[0]=0$

$f[i]=INF(1\le i\le n)$

代码

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#define LL long long
#define mem(a, b) memset(a, b, sizeof a)
#define lowbit(x) (-x&x)
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define endl '\n'
#define rev(x) reverse(x.begin(), x.end())
#define INF 0x3f3f3f3f

using namespace std;

const int N = 10010;

int w[N], v[N];
int n, m;
int f[N];

void solve() {
	int tt;
	cin >> tt;
	while (tt -- ) {
		int t1, t2;
		cin >> t1 >> t2;
		m = t2 - t1;
		
		cin >> n;
		for (int i = 1; i <= n; i ++ ) cin >> w[i] >> v[i];
		
		mem(f, 0x3f);
		
		f[0] = 0;
		for (int i = 1; i <= n; i ++ ) {
			for (int j = v[i]; j <= m; j ++ ) {
				f[j] = min(f[j], f[j - v[i]] + w[i]);
			}
		}
		
		if (f[m] != INF)
			printf("The minimum amount of money in the piggy-bank is %d.\n", f[m]);
		else puts("This is impossible.");
	}
}

int main() {
	IOS;
	
	solve();
	
	return 0;
}
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