NOIP 2006 普及组 开心的金明

二进制，超时

#include <bits/stdc++.h>
using namespace std;

const int M = 30;
int n, m, v[M], p[M], ans;	//n总钱数，m物品个数，v价格，p重要度 

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; i ++) scanf("%d%d", &v[i], &p[i]);
	
	int t = (1 << m);
	for (int i = 1; i < t; i ++) {
		int cost = 0, sum = 0;
		for (int j = 0; j < m; j ++) {
			if (i >> j & 1) {
				cost += v[j];
				if (cost > n) break;
				sum += v[j] * p[j];
			}
		}
		ans = max(ans, sum);
	}
	printf("%d", ans);

	return 0;
}

深搜

#include <bits/stdc++.h>
using namespace std;

const int M = 30;
int n, m, v[M], p[M], ans;	//n总钱数，m物品个数，v价格，p重要度 

//当前考虑第x个物品，前x-1个物品总价格是cost，乘积总和是sum 
void dfs(int x, int cost, int sum) {
	if (cost > n) return;
	if (x > m) {
		ans = max(ans, sum);
		return;
	}
	
	dfs(x + 1, cost + v[x], sum + v[x] * p[x]);
	dfs(x + 1, cost, sum);
}

int main() {
	cin >> n >> m;
	for (int i = 1; i <= m; i ++) cin >> v[i] >> p[i];
	
	dfs(1, 0, 0);
	cout << ans;

	return 0;
}

01背包

#include <bits/stdc++.h>
using namespace std;

const int N = 30009;
int n, m, v, p;	//n总钱数，m物品个数，v价格，p重要度 
int f[N];

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i ++) {
		scanf("%d%d", &v, &p);
		
		for (int j = n; j >= v; j --) {
			//f[j - v]表示在不买第i件物品的情况下，j - v元能够得到的乘积最大值 
			//买了第i件物品后，得到的乘积最大值是f[j - v] + v * p，花费变成了(j - v) + v 
			f[j] = max(f[j], f[j - v] + v * p);
		}
	}
	
	printf("%d", f[n]);

	return 0;
}