dp解法:
令dp[i]表示容量为i的背包所能得到的最大价值,考虑在当前物品集合中加入1个新考虑的物品i,则有如下状态转移方程:dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
#include <bits/stdc++.h> using namespace std; const int M = 1e4; typedef pair<int, vector<int> > piv; struct Node { int weight, value; int id; void read() { cin >> weight >> value; } void solve(piv dp[], int M) { for (int i = M; i >= weight; i --) { if (dp[i - weight].first + value > dp[i].first) { dp[i].first = dp[i - weight].first + value; dp[i].second = dp[i - weight].second; dp[i].second.push_back(id); } } } }; int n, m; piv dp[M]; int main() { puts("请输入背包容量和物品个数:"); cin >> m >> n; puts("请输入每个背包的重量(体积)和价值"); Node things; for (int i = 0; i < n; i ++) { things.read(); things.id = i + 1; things.solve(dp, m); } cout << "最大价值为:" << dp[m].first << endl; puts("选择的物品编号:"); for (int i = 0; i < dp[m].second.size(); i ++) { cout << dp[m].second[i] << " "; } cout << endl; return 0; }
贪心解法:
按部分背包的贪心策略,优先考虑单位价值高的物品,于是只需要按单位价值从高到低排序,然后依次考虑,能放则放即可。
#include <bits/stdc++.h> using namespace std; const int N = 1e2; struct Node { int weight, value; int id; void read() { cin >> weight >> value; } // 单位价值高的放前面 bool operator< (const Node &that) const { return value * that.weight > that.value * weight; } }; Node things[N]; int n, m; int main() { puts("请输入背包容量和物品个数:"); cin >> m >> n; puts("请输入每个背包的重量(体积)和价值"); for (int i = 0; i < n; i ++) { things[i].read(); things[i].id = i + 1; } sort(things, things + n); int ans = 0; vector<int> V; for (int i = 0; i < n; i ++) { if (m >= things[i].weight) { ans += things[i].value; m -= things[i].weight; V.push_back(things[i].id); } } cout << "最大价值为:" << ans << endl; puts("选择的物品编号为:"); for (int i = 0; i < V.size(); i ++) { cout << V[i] << " "; } cout << endl; return 0; }
时间: 2024-11-05 14:41:53