描述
Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W i and D i输出Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints样例输入
Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N(1 ≤ N≤ 3,402) available charms. Each charm iin the supplied list has a weight Wi(1 ≤ Wi≤ 400), a 'desirability' factor Di(1 ≤ Di≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M(1 ≤ M≤ 12,880).
Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.
Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W i and D i输出Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints样例输入
4 6 1 4 2 6 3 12 2 7样例输出
23
我的Runtime error代码
#include<iostream> #include<cstring> #include<vector> #include<algorithm> using namespace std; struct Node{ int w; int d; double dw; Node(){} }node[3410]; bool cp(Node a,Node b){ if(a.dw>b.dw) return true; else if(a.dw==b.dw&&a.w<b.w) return true; } int main(){ int n,m; cin>>n>>m; for(int i=0;i<n;i++){ cin>>node[i].w; cin>>node[i].d; node[i].dw=1.0*node[i].d/node[i].w; } sort(node,node+n,cp); int ct=0; for(int i=0;i<n;i++){ if(node[i].w<=m){ ct+=node[i].d; m-=node[i].w; }else continue; } cout<<ct; }
//意识到是dp了,但是还是不会写。
// 背包问题(动态规划) #include <iostream> #include <cstdio> #include <cstring> #define MAXN 3402 #define MAXM 12880 using namespace std; int main(){ int N, M, W[MAXN+5], D[MAXN+5], dp[MAXM+5]; while(scanf("%d%d", &N, &M) != EOF){ for(int i=0; i<N; i++){ scanf("%d%d", &W[i], &D[i]); } memset(dp, 0, sizeof(dp)); for(int i=0; i<N; i++){ for(int left_w=M; left_w>=W[i]; left_w--){ dp[left_w] = max(dp[left_w-W[i]]+D[i], dp[left_w]); } } printf("%d\n", dp[M]); } return 0; }PS:dp数组记录当容量为当前下标时,最高的价值;dp数组是动态覆盖的;每放一个物品就会更新一遍dp数组;当放第i个物品的时候,i-1及其之前的都已经计算好了。