【HDU - 2639】Bone Collector II （第K大背包，dp，STLset）

题干：

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 2 31).

Sample Input

Sample Output

12
2
0

题目大意：
01背包的第k个最大价值。注意，两种不同方法得到相同价值是算作同一个种。这意味着，可以得到的价值序列将是一个严格递减的序列，从第1个最大值，第2个最大值.....第k个最大值。
如果不同值的总数小于K，就输出"0"(不带引号)

解题报告：

直接用set维护前k大就可以了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
set<int> dp[1005];
int n,V,K;
int w[MAX],v[MAX];
int main()
{int T;cin>>T;while(T--) {cin>>n>>V>>K;for(int i = 0; i<=V; i++) dp[i].clear(),dp[i].insert(0);for(int i = 1; i<=n; i++) scanf("%d",v+i);for(int i = 1; i<=n; i++) scanf("%d",w+i);for(int i = 1; i<=n; i++) {for(int j = V; j>=w[i]; j--) {auto it = dp[j-w[i]].begin();for(;it!=dp[j-w[i]].end(); it++) {dp[j].insert(*it + v[i]);}while(dp[j].size() > K) dp[j].erase(dp[j].begin());}}printf("%d\n",*dp[V].begin());}return 0 ;
}

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