HDU 2602 Bone Collector【01背包入门题】
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Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 52370 Accepted Submission(s): 22079
Problem Description
Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
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, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. 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, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. 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 maximum of the total value (this number will be less than 231).
Sample Input
15 101 2 3 4 55 4 3 2 1
Sample Output
14
Author
Teddy
Source
HDU 1st “Vegetable-Birds Cup” Programming Open Contest
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lcy
01背包入门题:
转态转移方程:
f[i][v] = max ( f[i-1][v],f[i-1][v-c[i] ]+w[i] )
f[i][v]:前 i 件物品放入容量为v的背包获得最大价值。
c[i]: 第 i 件物品的体积。
w[i] :第 i 件物品的体积价值。
注意:输入数据的时候数组下标要从 1 开!
优化成一维的:
伪代码:
for i=1..N for v=V..0 f[v]=max{f[v],f[v-c[i]]+w[i]};
f[v] : 体积为v的背包的最大价值。
参考博客:http://www.wutianqi.com/?p=539
http://www.cnblogs.com/jiangjun/archive/2012/05/08/2489590.html
二维AC代码:
#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int maxn=1005;int v[maxn];int val[maxn];int dp[maxn][maxn];int main(){ int T,n,V; ios::sync_with_stdio(false); cin.tie(0); //freopen("data/2602.txt","r",stdin); cin>>T; while(T--) { cin>>n>>V; memset(dp,0,sizeof(dp)); //memset(v,0,sizeof(v)); //memset(val,0,sizeof(val)); for(int i=1;i<=n;i++) cin>>val[i]; for(int i=1;i<=n;i++) cin>>v[i]; for(int i=1;i<=n;i++) { for(int j=0;j<=V;j++) { if(j>=v[i]) dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+val[i]); else dp[i][j]=dp[i-1][j]; } } cout<<dp[n][V]<<endl; } return 0;}
优化成一维AC代码:
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){ int t; scanf("%d",&t); while(t--) { int n,V,v[1005],val[1005],dp[1005]; scanf("%d%d",&n,&V); int i,j,k; for(i=0; i<n; i++) scanf("%d",&val[i]); for(i=0; i<n; i++) scanf("%d",&v[i]); memset(dp,0,sizeof(dp)); for(i=0; i<n; i++) for(j=V; j>=v[i]; j--) dp[j]=max(dp[j],dp[j-v[i]]+val[i]); printf("%d\n",dp[V]); } return 0;}
尊重原创,转载请注明出处:http://blog.csdn.net/hurmishine
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