POJ 1190 生日蛋糕 剪枝

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 2147483648
#define N 33
#define MOD 1000000
#define INF 1000000009
const double eps = 1e-9;
const double PI = acos(-1.0);
/*
给定一个蛋糕的体积 和层数
求减去下底面后蛋糕表面积的最小值
从底层向上枚举，枚举每一层的可行H和R（加上之后小于V的）
当枚举到最后一层， 和最优解比较取值
*/
int V, M, ans = 0;//体积和层数
void DFS(int level, int preh, int prer, int sumv, int sumS)
{
    if (sumv > V)return;
    if (ans && sumS > ans) return;
    //cout << "R:: " << prer << "    H:: " << preh << endl;
    if (level > M)
    {
        /*if (sumv == V)
        {
            cout << "SUMV " << sumv << "  SUMS  " << sumS << endl;
        }*/
        if (sumv == V)
        {
            if (!ans || ans > sumS)
                ans = sumS;
        }
        return;
    }
    for (int r = M-level+1;r < prer; r++)
    {
        for (int h = M-level+1; h < preh; h++)
        {
            if (ans && sumS + r*2*h > ans)continue;
            if (sumv + (M - level + 1)*h*r*r < V ) continue;
            if (ans && sumS + 2 * (V - sumv) / r > ans) continue;//当前面积 加上剩余体积的最大面积 
            if (sumv + h*r*r <= V)
                DFS(level + 1, h, r, sumv + h*r*r, sumS + 2 * r*h);
        }
    }
}
int main()
{
    scanf("%d%d", &V, &M);
    for (int r = M; r <= 100; r++)
    {
        for (int h = M; h <= 1000; h++)
        {
            if (r*r*h*M < V|| r*r*h>V) continue;
            DFS(2, h, r, r*r*h, r*r + 2 * r*h);
        }
    }
    printf("%d\n", ans);
}