HDU 2602 Bone Collector【01背包入门题】
栏目：php教程时间：2016-12-05 13:56:45
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
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
Author
Teddy
Source
HDU 1st “Vegetable-Birds Cup” Programming
Open Contest
Recommend
lcy
01背包入门题：
转态转移方程：
f[i][v] = max ( f[i⑴][v],f[i⑴][v-c[i] ]+w[i] )
f[i][v]:前 i 件物品放入容量为v的背包取得最大价值。
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1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
2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[i⑴][j],dp[i⑴][j-v[i]]+val[i]);
else
dp[i][j]=dp[i⑴][j];
}
}
cout<<dp[n][V]<<endl;
}
return 0;
}
优化成1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
------分隔线----------------------------
------分隔线----------------------------