HDU 2191 【多重背包】

Input
输入数据首先包含一个正整数C，表示有C组测试用例，每组测试用例的第一行是两个整数n和m(1<=n<=100, 1<=m<=100),分别表示经费的金额和大米的种类，然后是m行数据，每行包含3个数p，h和c(1<=p<=20,1<=h<=200,1<=c<=20)，分别表示每袋的价格、每袋的重量以及对应种类大米的袋数。

Output
对于每组测试数据，请输出能够购买大米的最多重量，你可以假设经费买不光所有的大米，并且经费你可以不用完。每个实例的输出占一行。

Sample Input
1
8 2
2 100 4
4 100 2

Sample Output
400

【分析】：
【代码】：

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,n,x) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int MAXN =  1e3 + 5;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

int n,m,k;

int num[maxm];
int v[maxm], w[maxm],dp[maxm];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        ms(dp,0);
        cin>>m>>n;
        for(int i=0;i<n;i++){
            cin>>w[i]>>v[i]>>num[i];
        }
        for(int i=0;i<n;i++){
            for(int k=1; k<=num[i]; k++){
                for(int j=m; j>=w[i]; j--){
                    dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
                }
            }
        }
        cout<<dp[m]<<endl;
    }
    return 0;
}
/*
1
8 2
2 100 4
4 100 2


Sample Output
400
*/

【二进制优化】：

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,n,x) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int MAXN =  1e3 + 5;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

int n,m,k;

int num[maxm];
int v[maxm], w[maxm],dp[maxm];
void zero(int cost ,int value)
{
    for(int j=m;j>=cost;j--){
        dp[j]=max(dp[j],dp[j-cost]+value);
    }
}

void cmp(int cost, int value)
{
    for(int j=cost;j<=m;j++){
        dp[j]=max(dp[j],dp[j-cost]+value);
    }
}
void mul(int cost, int value,int cnt)
{
    if(m<=cnt*cost) {
        cmp(cost,value);
        return ;
    }
    else{
        int k=1;
        while(k<=cnt){
            zero(k*cost,k*value);
            cnt-=k;
            k<<=1;
        }
    }
    zero(cnt*cost,cnt*value);
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        ms(dp,0);
        cin>>m>>n;
        for(int i=1;i<=n;i++){
            cin>>w[i]>>v[i]>>num[i];
        }
        for(int i=1;i<=n;i++){
            mul(w[i],v[i],num[i]);
        }

        cout<<dp[m]<<endl;

    }
    return 0;
}
/*
1
8 2
2 100 4
4 100 2

400
*/