AtCoder Beginner Contest 286——D - Money in Hand

AtCoder Beginner Contest 286 题目讲解

A题 B题 C题 D题 E题

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蒟蒻来讲题，还望大家喜。若哪有问题，大家尽可提！

Hello, 大家好哇！本初中生蒟蒻今天以AtCoder Beginner Contest 286的D题——Money in Hand为例，给大家讲解一下判断图中存在闭环的常用方法！

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原题

Problem Statement

Takahashi has $N$ kinds of coins; specifically, for $1\le i\le N$, he has $B_i$​ coins worth $A_i$​ yen (the currency in Japan) each.

Determine if Takahashi can pay exactly $X$ yen (without change) with the coins he currently has.

Constraints

• $1\le N\le \le 50$
• $1\le X\le 10^4$
• $1\le A_i\le 100$
• $1\le B_i\le 50$
• $A_i$​ are pairwise distinct.
• All values in the input are integers.

Input

The input is given from Standard Input in the following format:

N  X
A1​  B1​
A2​  B2​
⋮
AN​  BN​

Output

Print Yes if Takahashi can pay exactly $X$ yen with the coins he currently has; print No otherwise.

Sample Input 1

2 19
2 3
5 6

Sample Output 1

Yes

Takahashi has three $2$-yen coins and six $5$-yen coins. He can use two $2$-yen coins and three $5$-yen coins to pay exactly $2\times 2+5\times 3=19$ yen. Thus, $Yes$ should be printed.

Sample Input 2

2 18
2 3
5 6

Sample Output 2

No

There is no combination of the coins that he can use to pay exactly $18$ yen. Thus, $No$ should be printed.

Sample Input 3

3 1001
1 1
2 1
100 10

Sample Output 3

Yes

He need not use all kinds of coins.

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思路

这道题真的挺简单，就是一个裸的恰好装满的多重背包问题，所以直接套模板即可。我们知道多重背包有三种解决方式，由于这道题数据比较少，三种都能用！（因为大家应该都会了，所以在此不一一讲解，若有不会的，私聊我，我一定竭尽所力的讲解）

三种代码

N为物品数量，V为物品最多的重量，S为物品的个数

第一种（时间复杂度：$O(N\times V\times S)$）

#include 

using namespace std;

const int M = 1e4 + 10;

int n, x;
int w[M], s[M];
bool dp[M];

int main()
{
	cin.tie(0);
	ios :: sync_with_stdio(0);
	
	cin >> n >> x;
	
	for (int i = 1; i <= n; i ++)
		cin >> w[i] >> s[i];
	
	dp[0] = 1;
	for (int i = 1; i <= n; i ++)
		for (int j = x; j >= 0; j --)
			for (int k = 1; k <= s[i] && k * w[i] <= j; k ++)
				dp[j] |= dp[j - w[i] * k];
				
	if (dp[x]) cout << "Yes" << endl;
	else cout << "No" << endl;
	
	return 0;
}

第二种二进制优化（时间复杂度：$O(N\times V\times log(S))$）

#include 
#include 
#include 

using namespace std;

const int N = 2e4 + 10;

int n, w;
int wt[N], vl[N], sz[N];
int f[N];

int main()
{
    cin >> n >> w;
    
    int ww, vv, ss;
    int cnt = 0;
    for (int i = 1; i <= n; i ++)
    {
        cin >> ww >> ss;
        
        int k = 1;
        while (k <= ss)
        {
            wt[++ cnt] = ww * k;
            ss -= k;
            k *= 2;
        }
        if (ss)
        {
            wt[++ cnt] = ww * ss;
        }
    }
    
    f[0] = 1;
    for (int i = 1; i <= cnt; i ++)
        for (int j = w; j >= wt[i]; j --)
            f[j] |= f[j - wt[i]];
            
    cout << ((f[w]) ? "Yes" : "No") << endl;
}

第三种单调队列优化（时间复杂度：$O(N\times V)$）

#include 
#include 
#include 

using namespace std;

const int N = 2e4 + 10;

int n, W;
int q[N], g[N], f[N];

int main()
{
    cin >> n >> W;
    
    f[0] = 1;
    for (int i = 0; i < n; i ++ )
    {
        int w, v, s;
        cin >> w >> s;
        
        memcpy(g, f, sizeof f);
        for (int j = 0; j < w; j ++)
        {
            int h = 0, t = -1;
            for (int k = j; k <= W; k += w)
            {
                if (h <= t && k - s * w > q[h]) h ++;
                
                while (h <= t && g[q[t]] <= g[k]) t --;
                
                if (h <= t) f[k] |= g[q[h]];
                q[++ t] = k;
            }
        }
    }
    
    (f[W]) ? puts("Yes") : puts("No");
}

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今天就到这里了！

大家有什么问题尽管提，我都会尽力回答的！最后，除夕夜祝大家新年快乐！

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