备战实习8-14

LCP 56. 信物传送

直接BFS

class Solution {
public:
    int conveyorBelt(vector<string>& matrix, vector<int>& start, vector<int>& end) {
        vector ach(matrix.size(),vector<int>(matrix[0].size(),(1<<30)));
        auto i = [&](){
            queue<pair<int,int>>q;
            int f[] = {0,1,0,-1,1,0,-1,0}; 
            string c = {">};
            q.push({start[0],start[1]});
            ach[start[0]][start[1]] = 0;
            while(!q.empty())
            {
                auto [x,y] = q.front();
                q.pop();
                for(int i=0;i<4;++i)
                {
                    int nx = x+f[i<<1];
                    int ny = y+f[i<<1|1];
                    if(nx>=0 && nx<matrix.size() && ny>=0 && ny<matrix[0].size())
                    {
                        if(ach[nx][ny]>ach[x][y]+(matrix[x][y]==c[i]?0:1))
                        {
                            ach[nx][ny] = ach[x][y]+(matrix[x][y]==c[i]?0:1);
                            q.push({nx,ny});
                        }
                    } 
                }
            }
            return ach[end[0]][end[1]];
        };
        return i();
    }
};

LCP 02. 分式化简

$a+\frac{1}{b}=\frac{a\ast b+1}{b}$

class Solution {
public:
    vector<int> fraction(vector<int>& cont) {
        int c = 1,d = 0;
        for(int i=cont.size()-1;i>=0;--i)
        {
            swap(c,d);
            c += cont[i]*d;
        }
        return {c/gcd(c,d),d/gcd(c,d)};
    }
};

LCP 04. 覆盖

状压dp,因为数据量较小，所以暴力处理每一行broken就行

class Solution {
public:
    int domino(int n, int m, vector<vector<int>>& broken) {
        vector<int>cl(n);
        vector<int>nums(n);
        vector br(n,vector<int>(m,0));
        for(auto i:broken)
        {
            br[i[0]][i[1]] = 1;
            cl[i[0]] |= (1<<i[1]);
        }
        vector dp(n,vector<int>((1<<m),0));
        for(int j=0;j<(1<<m);++j)
        {
            if(j&cl[0])
                continue;
            if(n>1 && (j&cl[1]))
                continue;
            for(int i=0;i<m-1;++i)
            {
                if(br[0][i]==0 && br[0][i+1]==0 && ((1<<i)&j)==0 && ((1<<(i+1))&j)==0)
                {
                    dp[0][j]++;
                    ++i;
                }
            }
            for(int i=0;i<m;++i)
            {
                if((1<<i)&j && n>1)
                    dp[0][j]++;
            }
        }
        for(int i=1;i<n;++i)
        {
            for(int j=0;j<(1<<m);++j)
            {
                if(j&cl[i])continue;
                if(i<n-1 && (j&cl[i+1]))
                    continue;
                for(int k=0;k<(1<<m);++k)
                    if(!(j&k))
                    {
                        int k1 = dp[i-1][k];
                        for(int z=0;z<m-1;++z)
                        {
                            if(br[i][z]==0 && br[i][z+1]==0 && ((1<<z)&j)==0 && ((1<<(z+1))&j)==0 && ((1<<z)&k)==0 && ((1<<(z+1))&k)==0)
                            {
                                k1++;
                                ++z;
                            }
                        }
                        if(i!=n-1)
                            for(int z=0;z<m;++z)
                                if((1<<z)&j)
                                    k1++;
                        dp[i][j] = max(k1,dp[i][j]);
                    }
            }
        }
        int ans = 0;
        for(int i=0;i<(1<<m);++i)
            ans = max(dp[n-1][i],ans);
        return ans;
    }
};

LCP 03. 机器人大冒险

计算循环一次的路径和终点(dx,dy)，对于路径每一个点(x1,y1)而言,如果有

$$\begin{gathered} x\%dx=x1 \\ y\%dy=y1 \\ x/dx=y/dy \end{gathered}$$

显然x/dx=y/dy恒成立，因为xdy=0,ydx=0
说明该路径会到达终点，同理障碍也是。

class Solution {
public:
    bool robot(string command, vector<vector<int>>& obstacles, int x, int y) {
        int dx=0,dy=0;
        for(auto i:command)
            if(i=='U')dy++;
            else dx++;
        int d1 = x/dx,d2 = y/dy;
        d1 = min(d1,d2);
        bool ok = 0;
        int nx=0,ny=0;
        if(d1*dx==x && d1*dy==y)
            ok = 1;
        for(auto i:command)
        {
            if(i=='U')ny++;
            else nx++;
            if(x-dx*d1==nx && y-dy*d1==ny)
                ok = 1;
            for(auto j:obstacles)
            {
                if(j[0]-min(j[0]/dx,j[1]/dy)*dx==nx &&  j[1]-min(j[0]/dx,j[1]/dy)*dy==ny && min(j[0]/dx,j[1]/dy)<=d1 && j[0] <=x &&j[1]<=y)
                    return 0;
                if(j[0]%dx==0 && j[1]%dy==0 && j[0]/dx==j[1]/dy && j[0] <=x &&j[1]<=y)
                    return 0;
            }
        }
        return ok;
    }
};