Bone Collector II
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5463 Accepted Submission(s):
2880
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.
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.
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
3 5 10 2 1 2 3 4 5 5 4 3 2 1 5 10 12 1 2 3 4 5 5 4 3 2 1 5 10 16 1 2 3 4 5 5 4 3 2 1
Sample Output
12 2 0
题目的意思就是求01背包的第k优解，则自然想到（我感觉一点都不自然）多一维，dp【j】【k】；
状态dp【j】的前k个最优解，都是由dp[j][1....k]和dp[j-w[i]][1.....k]+v[i]转移过来（没有证明过，但是对的），可以用优先队列来维护。
在求解dp[j][k]时，我们首先把dp[j][1....k]和dp[j-w[i]][1.....k]+v[i]统统放进优先队列（会自己从大到小排），然后我们依次拿出k个，放进dp[j][1.....k]就ok了，但是要避免重复。
#include <iostream> #include <string> #include <cstring> #include <algorithm> #include <queue> using namespace std; int main() { int T; int dp[1005][35]; cin >> T; priority_queue<int>q;//默认从大到小排 while (T--) { memset(dp, 0, sizeof(dp)); int n, vv, kk; cin >> n >> vv >> kk; int i, j, k; int v[105], w[105]; for (i = 1; i <= n; i++) cin >> v[i]; for (i = 1; i <= n; i++) cin >> w[i]; for (i = 1; i <= n; i++) { for (j = vv; j >= w[i]; j--)//01背包的循环 { while (!q.empty()) q.pop(); for (k = 1; k <= kk; k++) {//dp[j][1....k]和dp[j-w[i]][1.....k]+v[i]放进队列 q.push(dp[j][k]); q.push(dp[j - w[i]][k] + v[i]); } k = 1; while (1) { if (q.empty() || k == kk+1) break; if (k > 1 && q.top() != dp[j][k-1]) {//这一步避免重复， q.top() == dp[j][k-1]要排除 dp[j][k] = q.top(); k++; } else if (k == 1) { dp[j][k] = q.top(); k++; } q.pop(); } } } cout << dp[vv][kk] << endl; } return 0; }