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;
}