UVa12589 Learning Vector

大家都是用优雅的记忆化搜索做的，我最笨，懒得写记忆化，直接递推暴力转移了。

这题的难点在贪心上。自己想出来了

有人需要的话我就写详细证明吧。。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef pair<int, int> P;
#define x first
#define y second
const int MAXN = 50 + 2;

P v[MAXN];
int N, K;

inline bool cmp(const P &lhs, const P &rhs)
{
    return (double) ((double)lhs.y / (double)lhs.x) > 
    (double) ((double)rhs.y / (double)rhs.x);
}

int f[MAXN][MAXN][MAXN * MAXN];

int main()
{
    //freopen("12589.in", "r", stdin);
    int T;
    cin>>T;
    int cnt = 0;
    while(T--)
    {
        cnt++;
        cin>>N>>K;
        register int a, b;
        for(int i = 1; i <= N; i++)
        {
            scanf(" %d %d", &a, &b);
            v[i] = P(a, b);
        }
        
        sort(v + 1, v + N + 1, cmp);
        
        memset(f, -0x3f, sizeof(f));
        for(int i = 0; i <= N; i++)
            for(int j = 0; j <= K; j++)
                f[i][j][0] = 0;

        register int mxh[2] = {0}, d = 0;
        for(int i = 1; i <= N; i++)
        {
            d ^= 1;
            for(int j = 0; j <= min(i, K); j++)
                for(int h = 0; h <= mxh[d]; h++)
                {
                    f[i + 1][j + 1][h + v[i].y] = max(f[i + 1][j + 1][h + v[i].y], 
                    f[i][j][h] + (v[i].x * v[i].y) + 2 * (v[i].x * h));
                    f[i + 1][j][h] = max(f[i + 1][j][h], f[i][j][h]);
                    mxh[d ^ 1] = max(mxh[d ^ 1], h + v[i].y);
                }
        }

        int ans = 0;
        for(int i = 1; i <= max(mxh[0], mxh[1]); i++)
            ans = max(ans, f[N + 1][K][i]);
        printf("Case %d: %d
", cnt, ans);
    }
    return 0;
}