2016集训测试赛（十八）Problem C: 集串雷 既分数规划学习笔记

Solution

分数规划经典题. 话说我怎么老是忘记分数规划怎么做呀...
所以这里就大概写一下分数规划咯:
分数规划解决的是这样一类问题: 有(a_1, a_2 ... a_n)和(b_1, b_2 ... b_n)这样一些值(其中(b)严格大于零), 其中(a)和(b)之间存在某种联系; 要你决策出每个(a_k), 使得

[ans = frac{sum_{x = 1}^n a_x}{sum_{x = 1}^n b_x} ]

取得最大值或最小值.
考虑怎样解决: 考虑二分(ans), 找到一个最大的(ans)使得

[frac{sum_{x = 1}^n a_x}{sum_{x = 1}^n b_x} ge ans ]

稍作变换, 得到

[sum_{x = 1}^n a_x ge ans sum_{x = 1}^n b_x \ sum_{x = 1}^n a_x - b_x imes ans ge 0 ]

在每个位置(i)上决策(a)和(b)即可.
换到这一题, 直接照搬上述思路即可, 不再赘述了.

#include <cstdio>
#include <cctype>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;
namespace Zeonfai
{
    inline int getInt()
    {
        int a = 0, sgn = 1; char c;
        while(! isdigit(c = getchar())) if(c == '-') sgn *= -1;
        while(isdigit(c)) a = a * 10 + c - '0', c = getchar();
        return a * sgn;
    }
}
const int N = 500;
int suc[N + 1][2], a[N + 1], b[N + 1], sz[N + 1];
double f[N << 2][N + 2];
int main()
{

    #ifndef ONLINE_JUDGE

    freopen("zs.in", "r", stdin);
    freopen("zs.out", "w", stdout);

    #endif

    using namespace Zeonfai;
    int n = getInt(), cnt = n;
    for(int i = 1; i <= n; ++ i)
    {
        for(int j = 0; j < 2; ++ j)
        {
            _suc[i][j] = getInt();
            if(! _suc[i][j]) suc[i][j] = ++ cnt;
        }
        a[i] = getInt(); b[i] = getInt();
    }
    for(int i = cnt; i > n; -- i) sz[i] = 1;
    for(int i = n; i; -- i) sz[i] = sz[suc[i][0]] + sz[suc[i][1]];
    double L = 0, R = 10, ans;
    while(R - L > 1e-5)
    {
        double mid = (R + L) / 2;
        for(int i = 1; i <= n; ++ i) for(int j = 0; j <= n + 1; ++ j) f[i][j] = - 1e9;
        for(int i = cnt; i > n; -- i) for(int j = 0; j < 2; ++ j) f[i][j] = 0;
        for(int i = n; i; -- i) for(int j = 0; j <= sz[suc[i][0]]; ++ j) for(int k = 0; k <= sz[suc[i][1]]; ++ k)
            f[i][j + k] = max(f[i][j + k], f[suc[i][0]][j] + f[suc[i][1]][k] + abs(j - k) * a[i] - mid * (j ^ k ^ b[i]));
        int flg = 0;
        for(int i = 0; i <= sz[1]; ++ i) if(f[1][i] >= mid) flg = 1;
        if(flg) ans = mid, L = mid; else R = mid;
    }
    printf("%.2lf
", ans);
}