题解：AT_abc032_d [ABC032D] ナップサック問題

思路

subtask1

直接暴力搜索即可。

subtask2

普通的 01 背包，直接 \(dp\) 即可。

subtask3

改变 \(dp\) 的状态，设 \(dp_i\) 表示价值为 \(i\) 时用的最小体积，那么就直接在里面找最小值就行。

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;

typedef long long ll;
const int N = 200005;
ll n, W, v[N], w[N];
bool A = 1, B = 1;
ll dp[N];
ll ans;

void dfs(int x, ll sp, ll sum) {
	if (x > n) {
		ans = max(ans, sum);
		return ;
	}
	if (w[x] <= sp) 
		dfs(x + 1, sp - w[x], sum + v[x]);
	dfs(x + 1, sp, sum);
}

int main() {
	scanf("%lld%lld", &n, &W);
	for (int i = 1; i <= n; i++) {
		scanf("%lld%lld", &v[i], &w[i]);
		if (w[i] > 1000) A = 0;
		if (v[i] > 1000) B = 0;
	}	
	if (!A && !B) {
		dfs(1, W, 0);
		printf("%lld\n", ans);
	}
	else if (A) {
		for (int i = 1; i <= n; i++) {
			for (int j = W; j >= w[i]; j--) 
				dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
		}
		for (int i = 1; i <= W; i++)
			ans = max(ans, dp[i]);
		printf("%lld\n", ans);
	}
	else {
		memset(dp, 0x3f, sizeof(dp));
		dp[0] = 0;
		long long sumv = 0;
		for (int i = 1; i <= n; i++)
			sumv += v[i];
		for (int i = 1; i <= n; i++) {
			for (int j = 200000; j >= v[i]; j--) {
				dp[j] = min(dp[j], dp[j - v[i]] + w[i]);
			}
		}
		for (int i = 1; i <= 200000; i++)
			if (dp[i] <= W)
				ans = i;
		printf("%lld\n", ans);
	}
	return 0;
}