Google Code Jam 2014 Qualification 题解

拿下 ABD, 顺利晋级， 预赛的时候C没有仔细想，推荐C题，一个非常不错的构造题目！

A Magic Trick 简单的题目来取得集合的交并

#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
using namespace std;
int main()
{
    freopen("A-small-attempt0.in","r",stdin);
    freopen("A-small-attempt0.out","w",stdout);
    int T;
    cin>>T;
    for(int t=1; t<=T; t++)
    {
        int x,y;
        cin>>x;
        int m[4][4];
        set<int> X;
        for(int i=0; i<4; i++)
        {
            for(int j=0; j<4; j++)
            {
                cin>>m[i][j];
                if(i+1  == x) X.insert(m[i][j]);
            }
        }
        cin>>y;
        set<int> Y;
        for(int i=0; i<4; i++)
        {
            for(int j=0; j<4; j++)
            {
                cin>>m[i][j];
                if(i+1  == y) Y.insert(m[i][j]);
            }
        }
        vector<int> ret(X.size() + Y.size());
        auto itr = set_intersection(X.begin(),X.end(), Y.begin(),Y.end(), ret.begin());
        ret.resize(itr - ret.begin());
        cout<<"Case #"<<t<<": ";
        if(ret.size() == 1)
            cout<<ret[0]<<endl;
        else if(ret.size() > 1)
        {
            cout<<"Bad magician!"<<endl;
        } else  cout<<"Volunteer cheated!"<<endl;
    }
}

B Cookie Clicker Alpha  核心观察可以通过简单的推导获得一个upper  bound，然后枚举到这个upper bound就OK。

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
    freopen("B-large.in","r",stdin);
    freopen("B-large.out","w",stdout);
    int T;
    cin>>T;
    for(int t = 1; t<= T; t++)
    {
        double C,F,X;
        cin>>C>>F>>X;

        double ret = X/2.0;
        int up = max( (F*X - 2*C)/(F*C), 0.0);
        //cout<<up<<endl;
        double Nec = 0.0f;
        for(int k = 0; k< up; k++)
        {
            Nec += C/(2+k*F);
            ret = min(ret, Nec + X/(2+(k+1)*F));
        }
        printf("Case #%d: %.7f
", t, ret);
        //cout<<"Case #"<<t<<": "<<ret<<endl;
    }
}

C Minesweeper Master  这是一个构造的题目，非常非常的不错！核心观察在于扫雷的机制在边界的时候需要是2*n的这样一个结构才可能保证顺利完成边界情况。

#include <iostream>

using namespace std;

char Map[55][55];
int main()
{
    freopen("C-large-practice.in","r",stdin);
    freopen("C-large-practice.out","w",stdout);
    int T; cin>>T;

    for(int t= 1; t<=T; t++)
    {
        bool possible = false;
        int R,C,M;
        cin>>R>>C>>M;
        for(int i=0; i<R; i++)
        {
            for(int j=0; j<C; j++)
            {
                Map[i][j] = '*';
            }
        }
        if(R == 1 || C == 1 || R*C == M+1)
        {
            possible = true;
            Map[0][0] = 'c';
            int num = R*C - M-1;
            for(int i=0; i<R; i++)
            {
                for(int j=0; j<C; j++)
                {
                    if(i==0 && j==0) continue;
                    else if(num >0)
                    {
                        Map[i][j] = '.';
                        num--;
                    }else Map[i][j] = '*';
                }
            }
        }else
        {
            for(int r = 2; r<=R; r++)
            {
                for(int c = 2; c<=C; c++)
                {
                    int mineleft = M - (R*C - r*c);
                    if( mineleft <= (r-2)*(c-2) && mineleft >=0)
                    {
                        possible = true;
                        for(int i=0; i<R; i++) for(int j=0; j<C; j++)Map[i][j] = '*';
                        Map[0][0]= 'c';
                        for(int i=0; i<2; i++) for(int j=0; j<c; j++)
                        {
                            if(i ==0 && j==0) continue;
                            Map[i][j] = '.';
                        }
                        for(int i=2; i<r; i++) for(int j=0; j<2; j++) Map[i][j] = '.';
                        mineleft = (r-2)*(c-2) - mineleft;
                        for(int i= 2; i<r; i++) for(int j=2; j<c; j++)
                        {
                            if(mineleft > 0)
                            {
                                mineleft --;
                                Map[i][j] = '.';
                            } else Map[i][j] = '*';

                        }
                    }

                }
            }

        }
        cout<<"Case #"<<t<<":"<<endl;
        if(possible)
        {
            for(int i=0; i<R; i++)
            {
                for(int j=0; j<C; j++)
                {
                    cout<<Map[i][j];
                }
                cout<<endl;
            }
        }
        else
        {
            cout<<"Impossible"<<endl;
        }
    }
}

详细的分析参见：http://www.huangwenchao.com.cn/2014/04/gcj-2014-qual.html

D Deceitful War 类似于田忌赛马的问题，核心观察在于 每一个人的最优策略都是采用刚刚比你大的来进行比较。

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main()
{
    freopen("D-large.in","r",stdin);
    freopen("D-large.out","w",stdout);
    int T; cin>>T;
    for(int t=1; t<= T; t++)
    {
        int N;
        cin>>N;
        vector<double> A(N);
        vector<double> B(N);
        for(int i=0; i<N; i++) cin>>A[i];
        for(int i=0; i<N; i++) cin>>B[i];
        sort(A.begin(), A.end());
        sort(B.begin(), B.end());
        int War = 0;
        for(int i = A.size()-1, j= B.size()-1; i>=0 && j>= 0;)
        {
            if(B[j]>A[i])
            {
                War++;
                i--;
                j--;
            } else if(B[j] <A[i])
            {
                i--;
            }
        }
        //cout<<N - War<<endl;
        int NOWar = 0;
        for(int i = B.size()-1, j= A.size()-1; i>=0 && j>= 0;)
        {
            if(A[j]>B[i])
            {
                NOWar++;
                i--;
                j--;
            } else if(A[j] <B[i])
            {
                i--;
            }
        }
        //cout<<NOWar<<endl;
        cout<<"Case #"<<t<<": "<<NOWar<<" "<<N-War<<endl;
    }
    return 0;
}