hdu 2639 Bone Collector II(第K优背包)

最新推荐文章于 2021-01-18 14:47:36 发布

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最新推荐文章于 2021-01-18 14:47:36 发布

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分类专栏： DP 文章标签： integer each less struct output input

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Problem Description

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.

Input

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.

Output

One integer per line representing the K-th maximum of the total value (this number will be less than 2³¹).

Sample Input

Sample Output

背包问题求第K优解，在01背包的方程上加一维，即dp[i][v][k]。

这个方程可以理解为： dp[i][v]这个有序队列是由dp[i-1][v]和dp[i-1][v-c[i]]+w[i]这两个有序队列合并得到的。有序队列dp[i-1][v]即dp[i-1][v][1..K]，dp[i-1][v-c[i]]+w[i]则理解为在dp[i-1][v-c[i]][1..K]的每个数上加上w[i]后得到的有序队列。合并这两个有序队列并将结果（的前K项）储存到dp[i][v][1..K]中。

代码：

#include<stdio.h>
#include<string.h>

typedef struct node
{
	int vol, w;
}point;
point q[105];
int n, vol, k;
int dp[1005][35];
int x[35], y[35];

void zeroOnePack(point q)
{
	int a, b, c, j;
	for(int i=vol;i >=q.vol;i--)
	{
		for(j=0;j<k; j++)
		{
			x[j]=dp[i][j];
			y[j]=dp[i-q.vol][j]+q.w;
		}
		x[k]=y[k]=-1;  //结束标志
		a = b = c = 0;
		while(c<k && (a<k||b<k))
		{
			if(x[a]>y[b])
				dp[i][c]=x[a],a++;
			else
				dp[i][c]=y[b], b++;
			if(dp[i][c]!=dp[i][c-1])
				c++;
		}
	}
}

int main()
{
	int t;
	scanf("%d", &t);
	while(t--)
	{
                memset(dp, 0, sizeof(dp));
		scanf("%d %d %d", &n, &vol, &k);
		for(int i = 0; i < n; i++)
			scanf("%d", &q[i].w);
		for(int i = 0; i < n; i++)
			scanf("%d", &q[i].vol);
		for(int i = 0; i < n; i++)
			zeroOnePack(q[i]);
		printf("%d\n",dp[vol][k-1]);
	}
	return 0;
}