AOJ 275.Charm Bracelet

369

# 题目

Description  


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 i in 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.






Input  


  • Line 1: Two space-separated integers: N and M
  • Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di



Output  


  • Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints



Sample Input  


4 6
1 4
2 6
3 12
2 7


Sample Output  


23


题解



模板题,01背包问题


代码


/*
By:OhYee
Github:OhYee
HomePage:http://www.oyohyee.com
Email:oyohyee@oyohyee.com
 
かしこいかわいい?
エリーチカ!
要写出来Хорошо的代码哦~
*/
 
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
 
const int maxn = 20000;
int dp[maxn];
int v;
 
void ZeroOnePack(int cost,int weight) {
    for(int i = v; i >= cost; i--)
        dp[i] = max(dp[i],dp[i - cost] + weight);
}
 
bool Do() {
    int n,m;
    if(scanf("%d%d",&n,&m) == EOF)
        return false;
 
    v = m;
    memset(dp,0,sizeof(dp));
 
    for(int i = 1;i <= n;i++) {
        int a,b;
        scanf("%d%d",&a,&b);
        ZeroOnePack(a,b);
    }
 
    printf("%d\n",dp[m]);
    return true;
}
 
int main() {
    while(Do());
    return 0;
}
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