NKOJ2321 东方project

　　背包dp问题的变体，每一关看成一个背包，用的炸弹数看成重量，通关概率看成物品总价值，然后本关与之前所有关卡用的炸弹数最优分配用分类讨论。注意到若用100个炸弹则必定通关，那么枚举100个或剪枝都行。时间复杂度为o(100nm),最多正好是十的八次方。

 

#include<iostream>
#include<vector>
#include<cstdio>
using namespace std;
double dp[1005][1001];//dp[a][b]代表通a关用b个炸弹的最大通过概率

struct level {
    int k, b;
};

double min(double a, double b)
{
    return a < b ? a : b;
}

double max(double a, double b)
{
    return a > b ? a : b;
}

double mayPoint(int k, int b, int x)
{
    return min(1, double(k*x + b) / (double)100);
}


int main()
{
    int n, m;
    scanf("%d%d",&n,&m);
    vector<level> levels(n+1);
    for (int i = 1; i <= n; ++i)
    {
        int k, b;
        scanf("%d%d", &k, &b);
        levels[i].b = b;
        levels[i].k = k;
    }
    for (int i = 0; i <= m; ++i)
    {
        dp[0][i] = 1;
    }
    for (int i = 1; i <= n; ++i)
    {
        for (int nb = 0; nb <= m; ++nb)
        {
            for (int b = nb; b <= m; ++b)
            {
                dp[i][b] = max(dp[i][b], dp[i - 1][b - nb] * mayPoint(levels[i].k, levels[i].b, nb));
            }
            if (mayPoint(levels[i].k, levels[i].b, nb) >= 1)break;
        }
    }
    for (int i = 0; i <= n; ++i)
    {
        for (int j = 0; j <= m; ++j)
        {
            cout << dp[i][j] << "\t";
        }
        cout << endl;
    }
    printf("%lf\n", dp[n][m]);
    
}

（卡了一个小时最后还没做出来，菜swl）