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;
}