标签：this 入门 res time logs 石头 roo represent cond
Bone Collector II
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4948 Accepted Submission(s): 2588Problem DescriptionThe 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.
InputThe 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.
OutputOne integer per line representing the K-th maximum of the total value (this number will be less than 231).
Sample Input35 10 21 2 3 4 55 4 3 2 15 10 121 2 3 4 55 4 3 2 15 10 161 2 3 4 55 4 3 2 1Sample Output1220
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 using namespace std; 5 6 struct Node 7 { 8 int price; 9 int val; 10 } node[1005]; 11 12 int main() 13 { 14 int t; 15 scanf("%d",&t); 16 while(t--) 17 { 18 int n,v,k,i,dp[1005][31] = {0},a[31],b[31]; 19 scanf("%d%d%d",&n,&v,&k); 20 for(i = 0; i<n; i++) 21 scanf("%d",&node[i].price); 22 for(i = 0; i<n; i++) 23 scanf("%d",&node[i].val); 24 int j; 25 for(i = 0; i<n; i++) 26 { 27 for(j = v; j>=node[i].val; j--) 28 { 29 int cnt = 0,d; 30 for(d = 1; d<=k; d++)//分别将放入第i个石头与不放第i个石头的结果存入a,b,数组之中 31 { 32 a[d] = dp[j-node[i].val][d]+node[i].price; 33 b[d] = dp[j][d]; 34 } 35 int x,y,z; 36 x = y = z = 1; 37 a[d] = b[d] = -1; //边界！！！因为下面循环的时候会造成越过d-1； 38 while(z<=k && (x<=k || y<=k))//循环找出前K个的最优解 ，这个循环很关键啊 39 { 40 if(a[x] > b[y]) 41 { 42 dp[j][z] = a[x]; 43 x++; 44 } 45 else 46 { 47 dp[j][z] = b[y]; 48 y++; 49 } 50 if(dp[j][z]!=dp[j][z-1]) //判断有没有重复，重复就不计录下来 51 z++; 52 } 53 } 54 } 55 printf("%d\n",dp[v][k]); 56 } 57 58 return 0; 59 }
标签：this 入门 res time logs 石头 roo represent cond
原文地址：http://www.cnblogs.com/ISGuXing/p/7215341.html
