Bone Collector II

最新推荐文章于 2024-07-19 15:39:37 发布

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本文探讨了在给定背包容量和骨头数量的情况下，如何找到第K大骨头的总价值。通过动态规划方法，实现从最大价值到第K大价值的递减查找，解决了这一问题。

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题目描述

The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it doesn't matter,I will give you a link:

Here is the link: http://acm.hdu.edu.cn/showproblem.php?pid=2602

Today we are not desiring the maximum value of bones,but the K-th maximum value of the bones.NOTICE that,we considerate two ways that get the same value of bones are the same.That means,it will be a strictly decreasing sequence from the 1st maximum , 2nd maximum .. to the K-th maximum.

If the total number of different values is less than K,just ouput 0.

输入

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, K(N <= 100 , V <= 1000 , K <= 30)representing the number of bones and the volume of his bag and the K we need. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

输出

One integer per line representing the K-th maximum of the total value (this number will be less than 2 31).

样例输入

样例输出

12
2
0

题意：给出一行价值，一行体积，让你在v体积的范围内找出第k大的值.......（注意，不要和它的第一题混起来，它第一行是价值，再是体积）

思路：首先dp[i][j]代表的是在体积为i的时候第j优解为dp[i][j]......那么，我们就可以这样思考，i对应体积，那么如果只是一维的dp[i]，代表的应该是体积为i时的最大值，那么同理，dp[i][1]代表的是体积为i时的最大值，那么我们就可以退出两种动态，dp[i][m],dp[i-s[i][0]][m]+s[i][1].....然后把这两种状态开个两个数组分别保存起来，再合并出体积为i时的前k优解......依次后推，直到dp[v][k].......

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int dp[1005][35],ss[105][2];
int a[105],b[105];
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int n,m,k;
		scanf("%d%d%d",&n,&m,&k);
		for(int i=1;i<=n;i++)
		scanf("%d",&ss[i][0]);
		for(int i=1;i<=n;i++)
		scanf("%d",&ss[i][1]);
		if(k==0){
			printf("0\n");
			continue;
		}
		memset(dp,0,sizeof(dp));
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		for(int i=1;i<=n;i++){
			for(int j=m;j>=ss[i][1];j--){
				for(int l=1;l<=k;l++){
					a[l]=dp[j][l];
					b[l]=dp[j-ss[i][1]][l]+ss[i][0];
				}
				int x=1,y=1,w=1;
				a[k+1]=b[k+1]=-1;
				while(w<=k&&(x<=k||y<=k)){
					if(a[x]>b[y]) 
					dp[j][w]=a[x++];
					else
					dp[j][w]=b[y++];
					if(w==1||dp[j][w]!=dp[j][w-1])
					w++;
				}
			}
		}
		printf("%d\n",dp[m][k]);
	}
	return 0;
}