第一行两个整数V和m。
接下来m行,每行3个整数,表示第i种物品的数量、体积和价值。
,个数、体积、价值不超过1000。
输出一个整数,表示背包能装下的最大价值。
10 4 2 3 2 2 4 3 1 2 2 4 5 3
8
const rl = require("readline").createInterface({ input: process.stdin }); var iter = rl[Symbol.asyncIterator](); const readline = async () => (await iter.next()).value; void (async function () { // Write your code here const arr1 = (await readline()).split(" "); const capacity = parseInt(arr1[0]); const n = parseInt(arr1[1]); const amounts = []; const weights = []; const values = []; for (let i = 0; i < n; i++) { let tokens = (await readline()).split(" "); amounts.push(parseInt(tokens[0])); weights.push(parseInt(tokens[1])); values.push(parseInt(tokens[2])); } const dp = new Array(capacity + 1).fill(0); for (let i = 0; i < n; i++) { if (weights[i] >= capacity) { zero_oneBP(1); } let k = 1; while (k <= amounts[i]) { zero_oneBP(k); amounts[i] -= k; k = k * 2; } if (amounts[i] > 0) zero_oneBP(amounts[i]); function zero_oneBP(k) { for ( let remain_v = capacity; remain_v >= k * weights[i]; remain_v-- ) { let temp = k * values[i] + dp[remain_v - k * weights[i]]; dp[remain_v] = Math.max(temp, dp[remain_v]); } } } console.log(dp[capacity]); })();
import java.util.*; public class Main{ public static void main(String args[]){ Scanner sc = new Scanner(System.in); int v = sc.nextInt(); int m = sc.nextInt(); int []num = new int[m]; int []vArr = new int[m]; int []value = new int[m]; for(int i=0;i<m;i++){ num[i] = sc.nextInt(); vArr[i] = sc.nextInt(); value[i] = sc.nextInt(); } int [] dp = new int[v+1]; for(int i=0;i<m;i++){ if(num[i]*vArr[i]>=v){ for(int j=vArr[i];j<=v;j++){ dp[j] = Math.max(dp[j],dp[j-vArr[i]]+value[i]); } }else{ int k=1; int temp=num[i]; while(k<temp){ int v_tmp =k*vArr[i]; for(int j=v; j>=v_tmp;j--){ dp[j] = Math.max(dp[j],dp[j-v_tmp]+k*value[i]); } temp -=k; k*=2; } for(int j=v; j>=temp*vArr[i];j--){ dp[j] = Math.max(dp[j],dp[j-temp*vArr[i]]+temp*value[i]); } } } System.out.println(dp[v]); } }中间对数量进行了判断,当用不完的时候相当于完全背包,就用完全背包去解。毕竟完全背包这边不用拆分,时间上会短点。然后01背包用了二进制优化
w,n = map(int,input().split()) try: goods = [] while n: n-=1 goods.append(list(map(int,input().split()))) n = len(goods) new_goods= [] for i in range(n): for j in range(goods[i][0]): new_goods.append([goods[i][1],goods[i][2]])#体积,价值 #goods.clear() n = len(new_goods) #重新定义总数量 #01 背包 dp = [0]*(w+1) #第i个物品总容量为j时最大价值 for i in range(1,n+1): for j in range(w,new_goods[i-1][0]-1,-1): dp[j] = max(dp[j],dp[j-new_goods[i-1][0]]+new_goods[i-1][1]) print(dp[-1]) except: pass
/* 第一行两个整数V和m。 接下来m行,每行3个整数,表示第i种物品的数量、体积和价值。 V≤104, m≤500V\leq10^4,\ m≤500V≤104, m≤500,个数、体积、价值不超过1000。 */ #include<iostream> (720)#include<vector> using namespace std; int main() { int V, M; //体积为V 有m种物品 cin >> V >> M; vector<std::pair<int, int>> goods; //物品的体积 物品的价值 for (int i = 0; i < M; ++i) { int num, v, value; cin >> num >> v >> value; int n = 1; //物品分块的数量 while (n <= num) { // n应当小于1 goods.push_back({n * v, n * value}); num -= n; n <<= 1; } if (num != 0) goods.push_back({num * v, num * value}); } // dp[i][v]//选择第i次后 体积为v时的最大价值 vector<int> dp(V + 1, 0); for (int i = 1; i <= goods.size(); ++i) { for (int j = V; j > 0; --j) { if (j >= goods[i - 1].first) dp[j] = max(dp[j], dp[j - goods[i - 1].first] + goods[i - 1].second); } } cout << dp.back() << endl; return 0; }
#include <bits/stdc++.h> using namespace std; struct item { int w; int v; } l[10000]; int dp[10001]; int main() { int maxv, n; while (cin >> maxv >> n) { int cnt = 0; for (int i = 0; i < n; i++) { int v, w, k; cin >> k >> w >> v; int temp = 1; while (k - temp > 0) { k -= temp; l[++cnt].w = temp * w; l[cnt].v = temp * v; temp *= 2; } l[++cnt].w = w * k; l[cnt].v = v * k; } for (int i = 0; i <= maxv; i++) { dp[i] = 0; } for (int i = 1; i <= cnt; i++) { for (int j = maxv; j >= l[i].w; j--) { dp[j] = max(dp[j], dp[j - l[i].w] + l[i].v); } } cout << dp[maxv]; } return 0; }
#include<iostream> #include<vector> #include<algorithm> using namespace std; int V; void zeroonepack(vector<int> &result, vector<int> v, vector<int> value, int i) { for (int j = V; j >= v[i]; --j) j < v[i] ? result[j] = result[j] : result[j] = max(result[j], result[j - v[i]] + value[i]);//放不下第i个物体 return; } int main() { int n, temp1, temp2, temp3, k = 1;//背包体积V,n组数 cin >> V >> n; vector<int> num(n + 1, 0),v(1, 0), value(1, 0), result(V + 1, 0);/*v是物体体积,value是物体价值。result[i]表示前i个物品, 放到v中最大价值,需要初始化为0。如果需要刚好放满背包,除了第0个元素,其他要初始化为负无穷。v和value只申请一个是因为第0个要为0*/ for (int i = 1; i <= n; ++i) { cin >> temp1 >> temp2 >> temp3; num[i] = temp1; while (k < num[i])//未进入循环的时候,即等于原num[i]-2^k+1 { v.push_back(k*temp2); value.push_back(k*temp3); num[i] -= k; k = k * 2; } v.push_back(num[i] * temp2); value.push_back(num[i] * temp3);//补差值 k=1; } for (int i = 1; i <= v.size(); ++i)//二进制优化01背包后,有v.size()个物品 zeroonepack(result, v, value, i); cout << result[V] << endl; return 0; }
/*** @author: karlmical* @date: 20190805**/#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <string>#include <sstream>#include <string.h>using namespace std;const int MAXV = int(1e4) + 5;const int MAXN = 1000 + 1;int dp[MAXV];int capacity[MAXN], vol[MAXN], value[MAXN];void zeroone_knapsack(int limit, int i, int k){const int w = k * vol[i];for (int j = limit; j >= w; j--){int nv = dp[j - w] + k * value[i];if (nv > dp[j]){dp[j] = nv;}}}int multi_knapsack(int n, int limit){//initmemset(dp, 0, sizeof(int) * (n + 1));for (int i = 1; i <= n; i++){if (vol[i] * capacity[i] >= limit){for (int j = vol[i]; j <= limit; j++){int nv = dp[j - vol[i]] + value[i];if (nv > dp[j]){dp[j] = nv;}}}else{int k = 1, res = capacity[i];while (k < res){zeroone_knapsack(limit, i, k);res -= k;k <<= 1;}zeroone_knapsack(limit, i, res);}}return dp[limit];}int main(){int v, m;cin >> v >> m;for (int i = 1; i <= m; i++){cin >> capacity[i] >> vol[i] >> value[i];}cout << multi_knapsack(m, v);return 0;}
import java.util.Arrays; import java.util.Scanner; import java.util.ArrayList; /** * @Classname BeiBao_01 * @Description 有N中类型的商品,每一个有自己的体积w[i],价值v[i],容量是V,求最大的价值W * 注意:这是一个01背包问题,每一件商品只有1个,多重背包转换为01背包 * @Date 19-6-9 下午7:34 * @Created by mao<tianmao818@qq.com> */ public class Main { public static void main(String[] args){ Scanner sc=new Scanner(System.in); while (sc.hasNext()){ int W=sc.nextInt(); //容量 int N=sc.nextInt(); //代价 ArrayList<Integer> w=new ArrayList<>(); //价值 ArrayList<Integer> v=new ArrayList<>(); for(int i=0;i<N;i++){ //数量 int a=sc.nextInt(); //体积 int b=sc.nextInt(); //价值 int c=sc.nextInt(); for(int k=0;k<a;k++){ w.add(b); v.add(c); } } N=w.size(); int[] f=new int[W+1]; for (int i=0;i<N;i++){ for(int j=W;j>=w.get(i);j--){ //前i个商品对应不同容器的最大价值 f[j]=Math.max(f[j],f[j-w.get(i)]+v.get(i)); } } System.out.println(f[W]); } } }