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  • 单调队列优化DP,多重背包

     单调队列优化DP:http://www.cnblogs.com/ka200812/archive/2012/07/11/2585950.html

    单调队列优化多重背包:http://blog.csdn.net/flyinghearts/article/details/5898183

    传送门:hdu 3401 Trade

    /**************************************************************
        Problem:hdu 3401 Trade
        User: youmi
        Language: C++
        Result: Accepted
        Time:171MS
        Memory:17368K
    ****************************************************************/
    //#pragma comment(linker, "/STACK:1024000000,1024000000")
    //#include<bits/stdc++.h>
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <map>
    #include <stack>
    #include <set>
    #include <sstream>
    #include <cmath>
    #include <queue>
    #include <deque>
    #include <string>
    #include <vector>
    #define zeros(a) memset(a,0,sizeof(a))
    #define ones(a) memset(a,-1,sizeof(a))
    #define sc(a) scanf("%d",&a)
    #define sc2(a,b) scanf("%d%d",&a,&b)
    #define sc3(a,b,c) scanf("%d%d%d",&a,&b,&c)
    #define scs(a) scanf("%s",a)
    #define sclld(a) scanf("%I64d",&a)
    #define pt(a) printf("%d
    ",a)
    #define ptlld(a) printf("%I64d
    ",a)
    #define rep(i,from,to) for(int i=from;i<=to;i++)
    #define irep(i,to,from) for(int i=to;i>=from;i--)
    #define Max(a,b) ((a)>(b)?(a):(b))
    #define Min(a,b) ((a)<(b)?(a):(b))
    #define lson (step<<1)
    #define rson (lson+1)
    #define eps 1e-6
    #define oo 1e9
    #define TEST cout<<"*************************"<<endl
    const double pi=4*atan(1.0);
    
    using namespace std;
    typedef long long ll;
    template <class T> inline void read(T &n)
    {
        char c; int flag = 1;
        for (c = getchar(); !(c >= '0' && c <= '9' || c == '-'); c = getchar()); if (c == '-') flag = -1, n = 0; else n = c - '0';
        for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0'; n *= flag;
    }
    ll Pow(ll base, ll n, ll mo)
    {
        ll res=1;
        while(n)
        {
            if(n&1)
                res=res*base%mo;
            n>>=1;
            base=base*base%mo;
        }
        return res;
    }
    //***************************
    
    int T,P,w;
    const int maxn=2000+10;
    const ll mod=1000000007;
    int dp[maxn][maxn];
    int ap[maxn],bp[maxn];
    int as[maxn],bs[maxn];
    struct node
    {
        int val,p;
    };
    node q[maxn];
    int head,tail;
    int main()
    {
        #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        #endif
        int T_T;
        scanf("%d",&T_T);
        for(int kase=1;kase<=T_T;kase++)
        {
            sc3(T,P,w);
            rep(i,1,T)
            {
                sc2(ap[i],bp[i]);//buy,sell,money
                sc2(as[i],bs[i]);//buy,sell,number
            }
            rep(i,0,T)
                rep(j,0,P)
                    dp[i][j]=-oo;
            dp[0][0]=0;
            rep(i,1,T)
            {
                if(i<=w+1)
                {
                    rep(j,0,min(P,as[i]))
                        dp[i][j]=-ap[i]*j;
                }
                rep(j,0,P)
                    dp[i][j]=max(dp[i][j],dp[i-1][j]);
                if(i<=w+1)
                    continue;
                head=tail=0;
                for(int j=P;j>=0;j--)
                {
                    int temp=dp[i-w-1][j]+bp[i]*j;
                    while(tail>head&&temp>q[tail-1].val) tail--;
                    q[tail].val=temp,q[tail].p=j,tail++;
                    while(tail>head&&q[head].p-bs[i]>j)  head++;
                    temp=q[head].val;
                    dp[i][j]=max(dp[i][j],temp-bp[i]*j);
                }
                tail=head=0;
                rep(j,0,P)
                {
                    int temp=dp[i-w-1][j]+ap[i]*j;
                    while(tail>head&&temp>q[tail-1].val)  tail--;
                    q[tail].val=temp,q[tail].p=j,tail++;
                    while(tail>head&&q[head].p+as[i]<j) head++;
                    temp=q[head].val;
                    dp[i][j]=max(dp[i][j],temp-ap[i]*j);
                }
            }
            int ans=0;
            rep(i,0,P)
                ans=max(ans,dp[T][i]);
            pt(ans);
        }
        return 0;
    }
    View Code
    不为失败找借口,只为成功找方法
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  • 原文地址:https://www.cnblogs.com/youmi/p/5906475.html
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