hdu2639 Bone Collector II (01背包第k大解）

2019-11-08 01:38:39

字體：大中小

來源：轉載

供稿：網友

PRoblem 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=2602Today 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 Input

35 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 1 Sample Output1220 Authorteddy Source百萬秦關終屬楚 
題意：給出一行價值，一行體積，讓你在v體積的范圍內找出第k大的值；
思路：01背包的進一步優化，普通的背包是求得最大值；
          需要求第k大，我們需要在dp數組上多開一維，記錄v體積下第k大的值；
          然后我們利用01背包思想將原值與新得到的值進行按大小合并；
代碼：
#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int maxn=105;int dp[1010][50];int value[maxn],weight[maxn];int a[50],b[50];int n,V,K;void solve_dp(){    memset(dp,0,sizeof(dp));    for(int i=0;i<n;i++)        for(int j=V;j>=weight[i];j--)        {            for(int k=1;k<=K;k++)            {                a[k]=dp[j][k];                b[k]=dp[j-weight[i]][k]+value[i];            }            int x,y,z;            x=y=z=1;            a[K+1]=b[K+1]=-1;            while((z<=K)&&(x<=K||y<=K))  //合并            {                if(a[x]>b[y]) dp[j][z]=a[x++];                else dp[j][z]=b[y++];                if(dp[j][z]!=dp[j][z-1]) z++;            }        }}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d",&n,&V,&K);        for(int i=0;i<n;i++)            scanf("%d",&value[i]);        for(int i=0;i<n;i++)            scanf("%d",&weight[i]);        solve_dp();        printf("%d/n",dp[V][K]);    }}

上一篇：環境變量

下一篇：leetcode463