01 背包问题
2016-04-21 18:55
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http://acm.hdu.edu.cn/showproblem.php?pid=260201 背包问题的求解设大V为背包的容量,n为背包的个数v[i],w[i]分别代表第i个物品的体积和价值那么可得状态转移方程为dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+w[i])i代表前i个物品,j代表当前体积,即在前i个物品中选择几个物品放到容量为j的背包中能得到的最大价值。
例:5个物品 背包最大体积 6i=1;v[1]=5 w[1]=1i=2;v[2]=4 w[2]=2i=3;v[3]=3 w[3]=3i=4;v[4]=2 w[4]=4i=5;v[5]=1 w[5]=5 i j -- | -----1 2 3 4 5 6 1 0 0 0 0 1 1 2 0 0 0 2 2 2
3 0 0 3 3 3 3
4 0 4 4 4 7 7
5 5 5 9 9 9 12 dp[5][6]=12;优化:状态转移方程 :_dp[j]=max(_dp[j],_dp[j-v[i]]+w[i]) (j要逆序枚举)代码:#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>
#define maxn 10000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
bool cmp(int a,int b){
return a>b;
}
int n,V;
int v[1004],w[1003];
int dp[1004][1004];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int t;
cin>>t;
while(t--)
{
cin>>n>>V;
for(int i=1;i<=n;i++)
cin>>w[i];
for(int j=1;j<=n;j++)
cin>>v[j];
cle(dp);
///dp[i][j]=max(dp[i-1][j],dp[i][j-v[i]]+w[i]);
for(int i=1;i<=n;i++)
{
for(int j=0;j<v[i];j++)
dp[i][j]=dp[i-1][j];
for(int j=v[i];j<=V;j++)
dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+w[i]);
}
cout<<dp
[V]<<endl;
}
return 0;
}
空间优化代码:
例:5个物品 背包最大体积 6i=1;v[1]=5 w[1]=1i=2;v[2]=4 w[2]=2i=3;v[3]=3 w[3]=3i=4;v[4]=2 w[4]=4i=5;v[5]=1 w[5]=5 i j -- | -----1 2 3 4 5 6 1 0 0 0 0 1 1 2 0 0 0 2 2 2
3 0 0 3 3 3 3
4 0 4 4 4 7 7
5 5 5 9 9 9 12 dp[5][6]=12;优化:状态转移方程 :_dp[j]=max(_dp[j],_dp[j-v[i]]+w[i]) (j要逆序枚举)代码:#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>
#define maxn 10000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
bool cmp(int a,int b){
return a>b;
}
int n,V;
int v[1004],w[1003];
int dp[1004][1004];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int t;
cin>>t;
while(t--)
{
cin>>n>>V;
for(int i=1;i<=n;i++)
cin>>w[i];
for(int j=1;j<=n;j++)
cin>>v[j];
cle(dp);
///dp[i][j]=max(dp[i-1][j],dp[i][j-v[i]]+w[i]);
for(int i=1;i<=n;i++)
{
for(int j=0;j<v[i];j++)
dp[i][j]=dp[i-1][j];
for(int j=v[i];j<=V;j++)
dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+w[i]);
}
cout<<dp
[V]<<endl;
}
return 0;
}
空间优化代码:
#include<iostream> #include<cstring> #include<algorithm> #include<cstdlib> #include<vector> #include<cmath> #include<stdlib.h> #include<iomanip> #include<list> #include<deque> #include<map> #include <stdio.h> #include <queue> #define maxn 10000+5 #define ull unsigned long long #define ll long long #define reP(i,n) for(i=1;i<=n;i++) #define rep(i,n) for(i=0;i<n;i++) #define cle(a) memset(a,0,sizeof(a)) #define mod 90001 #define PI 3.141592657 const ull inf = 1LL << 61; const double eps=1e-5; using namespace std; bool cmp(int a,int b){ return a>b; } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int t;cin>>t; int n,v;int value[1002],vol[1002];int dp[1002]; while(t--) { cle(dp); cin>>n>>v; for(int i=1;i<=n;i++) cin>>value[i]; for(int i=1;i<=n;i++) cin>>vol[i]; for(int j=1;j<=n;j++) for(int k=v;k>=vol[j];k--) dp[k]=max(dp[k],dp[k-vol[j]]+value[j]); cout<<dp[v]<<endl; } return 0; }对于空间优化的代码可以理解为,空一次遍历,这样才能使dp[j-v[i]]等价于dp[i-1][j-v[i]]。看图
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