NOIP-贪心，递推，枚举，模拟 分类： noip bzoj hiho 2015-08-04 08:31 8人阅读 评论(0) 收藏

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bzoj2241

枚举矩形的长和宽，判断是否可行。时间复杂度：O((n*m)^3)
剪枝：1.最优化剪枝 2.如果矩形面积不能整除地图点权之和，则直接判断不可行。

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const int maxn = 105, maxm = 105;
int n, m, sum, ans;
int map[maxn][maxm];
int tmp[maxn][maxm];

bool check(int row,int col)
{
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            tmp[i][j] = map[i][j];

    for(int i = 1; i + row - 1 <= n; i++)
        for(int j = 1; j + col - 1 <= m; j++)
            if(tmp[i][j])
            {
                for(int p = row - 1; p >= 0; p--)
                    for(int q = col - 1; q >= 0; q--)
                        tmp[i+p][j+q] -= tmp[i][j]; 
            }
     for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            if(tmp[i][j]) return false;     

     return true;       
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj2241.in","r",stdin);
    freopen("bzoj2241.out","w",stdout);
#endif

    read(n), read(m);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            read(map[i][j]), sum += map[i][j];

    for(int i = n; i > 0; i--)
        for(int j = m; j > 0; j--)
            if(i*j > ans && !(sum%(i*j)) && check(i,j)) ans = i*j;

    write(sum/ans);

#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
#endif
    return 0;       
}

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uva11729

按执行时间排序，贪心即可。

证明：交换相邻安排任务的顺序，证明一定不比原决策更优。

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int main()
{
    int CaseNum = 0;

#ifndef ONLINE_JUDGE
    freopen("Uva11729.in","r",stdin);
    freopen("Uva11729.out","w",stdout);
#endif

    while(scanf("%d",&n) != EOF)
    {
        if(!n) break;

        ++CaseNum, ans = 0;

        printf("Case %d: ",CaseNum);

        for(int i = 1; i <= n; i++)
            std::cin >> c[i].second >> c[i].first;

        std::sort(c + 1, c + n + 1);

        for(int i = n, t = 0; i > 0; i--)
            t += c[i].second ,ans = std::max(ans, t + c[i].first);

        write(ans), puts("");
    }

#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
#endif
    return 0;       
}

---

hiho 1162

可以找规律，OEIS上有定义: http://oeis.org/A001835

Number of ways of packing a 3 X 2(n-1) rectangle with dominoes. - David Singmaster.

f(1)=1   f(2)=3   f(n)=4∗f(n−1)−f(n−2)(n>2)

状态压缩的递推也可以处理，当然两种递推都需要矩阵乘法优化。

---

const int Mod = 12357;

int n;

struct Matrix
{
    long long a[2][2];
}P, T, I, emp;
Matrix operator *(const Matrix &A,const Matrix &B)
{
    Matrix ret = emp;

    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++)
        {
            for(int k = 0; k < 2; k++)
                ret.a[i][j] += A.a[i][k]*B.a[k][j]%Mod;
            ret.a[i][j] %= Mod;     
        }
    return ret; 
}
Matrix powermod(Matrix Q,int k)
{
    Matrix ret = I;
    while(k)
    {
        if(k&1) ret = ret*Q;
        Q = Q*Q, k >>= 1;
    }
    return ret;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("cover.in","r",stdin);
    freopen("cover.out","w",stdout);
#endif

    read(n);

    if(n&1) putchar('0');
    else
    {
        n >>= 1;

        T.a[0][0] = 4;
        T.a[0][1] = 1;
        T.a[1][0] = Mod - 1;
        T.a[1][1] = 0;

        P.a[0][0] = 3;
        P.a[0][1] = 1;

        I.a[0][0] = 1;
        I.a[1][1] = 1;

        P = P*powermod(T, n - 1);

        write(P.a[0][0]);
    }

#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
#endif
    return 0;       
}

---

NOIP2012 国王游戏

贪心按 ai∗bi 排序，然后高精度处理。

---

const int maxn = 1005, Base = 10, maxl = 10000;
int n, a[maxn], b[maxn];
std::pair<int,int> c[maxn];
BigNum emp, poi, tmp, ans;

int main()
{
#ifndef ONLINE_JUDGE
    freopen("game.in","r",stdin);
    freopen("game.out","w",stdout);
#endif

    read(n), read(a[0]), read(b[0]);

    for(int i = 1; i <= n; i++)
        read(a[i]), read(b[i]), c[i] = std::make_pair(a[i]*b[i], b[i]);

    std::sort(c + 1, c + n + 1);    

    for(int i = 1; i <= n; i++)
        a[i] = c[i].first/c[i].second, b[i] = c[i].second;

    poi.len = 1, poi.num[1] = 1;
    for(int i = 1; i <= n; i++)
    {
        poi *= a[i-1], tmp = poi, tmp /= b[i];

        if(tmp > ans) ans = tmp;
    }

    ans.print();

#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
#endif
    return 0;       
}

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