• POJ


    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
    ",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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  • 原文地址:https://www.cnblogs.com/fht-litost/p/9207049.html
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