Bone Collector II
Time Limit: 2000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 263964-bit integer IO format: %I64d Java class name: Main
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
Source
解题：dp背包。。。每次都排下序。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #include <stack> 13 #define LL long long 14 #define pii pair<int,int> 15 #define INF 0x3f3f3f3f 16 using namespace std; 17 const int maxn = 1005; 18 int dp[maxn][75],v[maxn],c[maxn],tmp[maxn*10]; 19 int N,V,K; 20 int main() { 21 int t,cnt,z; 22 scanf("%d",&t); 23 while(t--){ 24 scanf("%d %d %d",&N,&V,&K); 25 for(int i = 1; i <= N; i++) 26 scanf("%d",v+i); 27 for(int i = 1; i <= N; i++) 28 scanf("%d",c+i); 29 memset(dp,0,sizeof(dp)); 30 for(int i = 1; i <= N; i++){ 31 for(int j = V; j >= c[i]; j--){ 32 cnt = 0; 33 for(int k = 1; k <= K; k++){ 34 tmp[cnt++] = dp[j][k]; 35 tmp[cnt++] = dp[j-c[i]][k]+v[i]; 36 } 37 sort(tmp,tmp+cnt); 38 z = 1; 39 for(int i = cnt-1; i >= 0; i--){ 40 if(z > K) break; 41 if(i == cnt-1 || tmp[i] != tmp[i+1]) 42 dp[j][z++] = tmp[i]; 43 } 44 } 45 } 46 printf("%d ",dp[V][K]); 47 } 48 return 0; 49 }