HDOJ 4341 Gold miner（分组DP）

就题目来说还是属于比较常规的分组DP，处理点与点之间的关系稍微麻烦点，借鉴了网上结构体重载的方法

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <algorithm>



using namespace std;

#define max(a, b)  ((a) > (b) ? (a) : (b))



const int MAXN = 205;

int dp[40005];



struct Point {

    int x, y;

    int t, w;

    void input() {

        scanf("%d %d %d %d", &x, &y, &t, &w);

    }

    int operator * (const Point &p) const {

        return x * p.y - y * p.x;

    }

    friend int dis(const Point &x) {

        return x.x * x.x + x.y * x.y;

    }

    friend bool operator < (const Point &a, const Point &b) {

        int temp = a * b;

        if (temp == 0)

            return dis(a) < dis(b);

        return temp < 0;

    }

} p[MAXN];



int main()

{

    int n, time;

    int c = 0;

    while (scanf("%d %d", &n, &time) != EOF)

    {

        for (int i = 0; i < n; ++i)

            p[i].input();

        sort(p, p + n);



        memset(dp, 0, sizeof(dp));

        //dp[0] = 0;

        for (int i = 0; i < n; ++i)

        {

            int j = i;

            while (j + 1 < n && p[j+1] * p[i] == 0)

                ++j;

            

            int tt, ww;

            for (int t = time; t > 0; --t)

            {

                tt = 0; ww = 0;

                for (int k = i; k <= j; ++k)

                {

                    tt += p[k].t;

                    ww += p[k].w;

                    if (t >= tt)

                        dp[t] = max(dp[t], dp[t-tt] + ww);

                }

            }

            i = j;

        }

        ++c;

        printf("Case %d: %d\n", c, dp[time]);

    }

    return 0;

}