POJ - 1276 Cash Machine(多重背包)

https://vjudge.net/problem/POJ-1276

题意

给一个最大额度cash,现有N种面值的纸币,第i种纸币有ni张,价值为Di。问用这些纸币能凑出来的最大金额数,要求不超过cash。

分析

多重背包问题。这里的Di即是价值也是花费。模板题

#include<iostream>
#include<cmath>
#include<cstring>
#include<queue>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<map>
#include<set>
#define rep(i,e) for(int i=0;i<(e);i++)
#define rep1(i,e) for(int i=1;i<=(e);i++)
#define repx(i,x,e) for(int i=(x);i<=(e);i++)
#define X first
#define Y second
#define PB push_back
#define MP make_pair
#define mset(var,val) memset(var,val,sizeof(var))
#define scd(a) scanf("%d",&a)
#define scdd(a,b) scanf("%d%d",&a,&b)
#define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define pd(a) printf("%d\n",a)
#define scl(a) scanf("%lld",&a)
#define scll(a,b) scanf("%lld%lld",&a,&b)
#define sclll(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)
#define IOS ios::sync_with_stdio(false);cin.tie(0)

using namespace std;
typedef long long ll;
template <class T>
void test(T a){cout<<a<<endl;}
template <class T,class T2>
void test(T a,T2 b){cout<<a<<" "<<b<<endl;}
template <class T,class T2,class T3>
void test(T a,T2 b,T3 c){cout<<a<<" "<<b<<" "<<c<<endl;}
template <class T>
inline bool scan_d(T &ret){
    char c;int sgn;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret = ret*10+(c-'0');
    ret*=sgn;
    return 1;
}
//const int N = 1e6+10;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const ll mod = 1000000000;
int T;

void testcase(){
    printf("Case %d:",++T);
}

const int MAXN = 5e5+10 ;
const int MAXM = 250;
const double eps = 1e-8;
const double PI = acos(-1.0);

int N,V;
int dp[100005];
int n[15],d[15];
void ZeroOnePack(int c,int w){
    for(int i=V;i>=c;i--){
        dp[i]=max(dp[i],dp[i-c]+w);
    }
}

void CompletePack(int c,int w){
    for(int i=c;i<=V;i++){
        dp[i]=max(dp[i],dp[i-c]+w);
    }
}

void MultiplePack(int c,int w,int m){
    if(c*m>=V){
        CompletePack(c,w);
    }else{
        int k=1;
        while(k<m){
            ZeroOnePack(k*c,k*w);
            m=m-k;
            k<<=1;
        }
        ZeroOnePack(m*c,m*w);
    }
}

int main() {
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif // LOCAL
    while(~scanf("%d",&V)){
        scanf("%d",&N);
        for(int i=1;i<=N;i++){
            scanf("%d%d",&n[i],&d[i]);
        }
        if(N==0 || V==0){
            puts("0");
            continue;
        }
        mset(dp,0);
        for(int i=1;i<=N;i++){
            MultiplePack(d[i],d[i],n[i]);
        }
        cout<<dp[V]<<endl;
    }
    return 0;
}

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转载自www.cnblogs.com/fht-litost/p/9207049.html