[Pregunta solución] CF922E: Aves

Portal del título original
Este es un problema de mochila
$d{p}_{yo}$Indica un total de $El$ maná máximo restante para$i$ pájaro.
Cada vez que camino por un árbol, todos los$d$ valor$p$$+X$
trata al pájaro como un artículo cada vez, la división binaria hace 01 mochila

Código:

#include <bits/stdc++.h>
#define maxn 1010
#define maxm 10010
#define LL long long
using namespace std;
LL W, B, X, cost[maxn], power[25], c[maxn], dp[maxm], sum;
int n;

inline int read(){
    
    
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

int main(){
    
    
	n = read(), W = read(), B = read(), X = read();
	for (int i = 1; i <= n; ++i) c[i] = read();
	for (int i = 1; i <= n; ++i) cost[i] = read();
	for (int i = 0; i <= 10000; ++i) dp[i] = -1;
	dp[0] = W;
	power[0] = 1;
	for (int i = 1; i <= 20; ++i) power[i] = power[i - 1] << 1;
	for (int i = 1; i <= n; ++i){
    
    
		sum += c[i];
		for (int j = 0; j <= sum; ++j)
			if (dp[j] != -1) dp[j] = min(dp[j] + X, W + 1LL * B * j);
		for (int j = 0; j <= 14; ++j)
			if (c[i] >= power[j]){
    
    
				c[i] -= power[j];
				for (int k = sum; k >= power[j]; --k){
    
    
					if (dp[k - power[j]] == -1) continue;
					if (dp[k - power[j]] - power[j] * cost[i] < 0) continue;
					dp[k] = max(dp[k], dp[k - power[j]] - power[j] * cost[i]);
				}
			}
		if (c[i])
			for (int k = sum; k >= c[i]; --k){
    
    
				if (dp[k - c[i]] == -1) continue;
				if (dp[k - c[i]] - c[i] * cost[i] < 0) continue;
				dp[k] = max(dp[k], dp[k - c[i]] - c[i] * cost[i]);
			}
	}
	for (int i = sum; i >= 0; --i)
		if (dp[i] != -1){
    
     printf("%d\n", i); break; }
	return 0;
}