uva-757-贪心

题意：有个人要去湖里钓鱼,总共有N个湖,排成一个序列,用字母P表示湖，从湖pi 到 pi+1(下一个湖)需要ti个五分钟.

并且，每个湖里可钓出来的鱼每过五分钟就减少di.如果产出的鱼小于等于di.那么下个五分钟这个湖再也不会产出鱼.

问:使得总的钓出鱼数目最大和每个湖花费的对应时间,.如果鱼数目最大存在多种解,让时间尽可能花费在前面的湖上.

解法：

枚举每一个pi,划去从p0至pi的总time(总的转移时间).

这样我们就可以在任意p0-pi之间钓鱼而不用考虑time的问题.剩下的h全部花费在钓鱼上.
每次花费的五分钟都花在产出鱼数目最大的湖上.这样得到总鱼数目总是最大的.如果所有的鱼都钓完还有时间剩余.全部加在第一个湖上面.

#include <string>
#include<iostream>
#include<map>
#include<memory.h>
#include<vector>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<math.h>
#include<iomanip>
#include<bitset>
#include"math.h"
namespace cc
{
    using std::cout;
    using std::endl;
    using std::cin;
    using std::map;
    using std::vector;
    using std::string;
    using std::sort;
    using std::priority_queue;
    using std::greater;
    using std::vector;
    using std::swap;
    using std::stack;
    using std::queue;
    using std::bitset;

    class  State
    {
    public:

        int curFish;
        int d;
        int time;
        int index;
        State() {}
        State(int curFish, int d, int time, int index) :
            curFish(curFish), d(d), time(time), index(index)
        {

        }
        friend bool operator < (const State&  s1, const State&  s2)
        {
            if (s1.curFish != s2.curFish)
                return s1.curFish < s2.curFish;
            return s1.index > s2.index;
        }
    };

    constexpr int N = ;
    int H = ;
    int n;

    int totalFish;
    int per[N] = {  };
    State states[N];

    priority_queue < State>q;

    void solve()
    {
        int t = ;
        while (cin >> n && n)
        {
            if (t != )
                cout << endl;
            ++t;
            cin >> H;
            H = H * ;
            int f = ;
            for (int i = ;i < n;i++)
            {
                cin >> f;
                State state(f, , , i);
                states[i] = state;
            }
            for (int i = ;i < n;i++)
            {
                cin >> states[i].d;
            }
            for (int i = ;i < n;i++)
            {
                cin >> states[i].time;
                states[i].time = states[i].time + states[i-].time;
            }
            totalFish = -;
            memset(per, , sizeof(per));
            for (int i = ;i < n;i++)
            {
                while (q.empty() == false)
                    q.pop();
                for (int k = ;k <= i;k++)
                    q.push(states[k]);
                int h = H;
                int curTotalFish = ;
                int times[N] = {};
                h = h - states[i].time;
                while (h > )
                {
                    State s = q.top();
                    q.pop();
                    int f = s.curFish;
                    if (f <= )break;
                    --h;
                    s.curFish -= s.d;
                    curTotalFish += f;
                    times[s.index]++;
                    q.push(s);
                }
                if (h > )
                    times[] += h;
                if (curTotalFish > totalFish)
                {
                    totalFish = curTotalFish;
                    memcpy(per, times, sizeof(times));
                }
            }
            cout << per[] * ;
            for (int i=;i<n;i++)
                cout<<", " << per[i] * ;
            cout << endl;
            cout << "Number of fish expected: " << totalFish << endl;
        }
    }
};

int main()
{

#ifndef ONLINE_JUDGE
    freopen("d://1.text", "r", stdin);
#endif // !ONLINE_JUDGE
    cc::solve();

    return ;
}