算法竞赛入门经典第三章总结(2):后半部分习题解答

UVA232:题目就不写了,主要是模拟单词的变换.注意必须要按照启示格的顺序来输出.在输出竖的单词时候要用一个矩阵来标记是否被访问过.当然还可以按照竖的方式写入然后再排序.

def cross(matrix,r,c):
    mark = [[0]*c for i in range(r)]
    def test(i,j):#检查一个白格是不是启示格
        #if(matrix[i][j]=='U'):print(mark)
        if(i==0 or j==0):return True
        elif(matrix[i-1][j] == '*' or matrix[i][j-1]=='*'):return True
        else:return False
    i = 0;across = ''
    while(i<r):
        j = 0
        while(j<c):
            #print(matrix[i][j])
            if(matrix[i][j]!='*' and test(i,j)):
                across = ''
                across+=matrix[i][j]
                while(j+1<c and matrix[i][j+1]!='*'):across+=matrix[i][j+1];j+=1
                j+=1
                print(across)
            else: j+=1
        i+=1
    print('Down:')
    i = 0;across = ''
    while(i<r):
        j = 0
        while(j<c):
            if(matrix[i][j]!='*' and test(i,j) and mark[i][j]!=1):
                across = ''
                mark[i][j] = 1#标记已被访问过
                across+=matrix[i][j]
                _i = i;
                while(_i+1<r and matrix[_i+1][j]!='*'):
                    #print(_i+1,j)
                    across+=matrix[_i+1][j]
                    mark[_i+1][j] = 1
                    _i+=1
                j+=1
                print(across)
            else: j+=1
        i+=1

UVA1368:DNA序列

只需要一列一列的进行比较就可以了,因为每一个位置的DNA都是独立的,与其它DNA无关

def Hamming(buf,m,n):
    ans = ''
    diff = 0
    for j in range(n):
        count = {'A':0,'C':0,'G':0,'T':0}
        max = 0
        for i in range(m):
            count[buf[i][j]]+=1
            if(count[buf[i][j]] > max):
                max = count[buf[i][j]]
                max_alpha = buf[i][j]
            elif(count[buf[i][j]] == max):#比较字典序
                if(buf[i][j]<max_alpha):
                    max = count[buf[i][j]]
                max_alpha = buf[i][j]
        ans += max_alpha
        diff += (m - max)
    print(ans,diff)
Hamming(['TATGATAC','TAAGCTAC','AAAGATCC','TGAGATAC','TAAGATGT'],5,8)

UVA202,这个题目的关键不是去判断小数,而是判断每一次的余数,只要余数重复那么小数就会重复的.想到这一点就不难了.

def Repeating(a,b):
    ans1 = a//b
    a = a%b
    tmp = ''
    ans2 = str((a*10)//b)
    mark_store = []
    while(True):
        a = (a*10)%b
        tmp = (a*10)//b
        if(a in mark_store):break
        else:
            mark_store.append(a)
            ans2 += str(tmp)
    m = ans2[:-1]
    print(str(ans1)+'.'+'(%s)'%(m),len(m))

UVA10340:这个题目关键是要逆着思考,考虑s字符串中的字符是否可以在t中按顺序找到,即只能向后查找.这样就能很容易的判断了.O(m+n)的复杂度。

def Allinall(s,t):#s-->t??
    index = -1
    for ch in s:
        index = t.find(ch,index+1)
        if(index == -1):print('NO');return False
    print('YES')
Allinall('dc','abcde')

UVA1587,此题关键是想到将输入标准化,统计边的情况之后进行检验.一定满足3种的分布.3条不同的边,2条相同,3条都一样.

def Box(L,W):
    def mysort():
        for i in range(6):
            if(L[i]<W[i]):L[i],W[i] = W[i],L[i]
    mysort()
    X = [(L[i],W[i]) for i in range(6)]
    Count = {}
    for each in X:
        if str(each) not in Count:
            Count[str(each)] = 1
        else:Count[str(each)] += 1
    if(len(Count)==3):
        print(Count)
        for each,val in Count.items():
            if(val != 2):print('Impossible 4');return
        print('Possible')
    elif(len(Count)==2):
        test = [x for k,x in Count.items()]
        test.sort()
        if (test!=[2,4]):print('Impossible 8');return
        print('Possible')
    elif(len(Count)==1):
        for each,val in Count.items():
            if(Count[each]!=6):print('Impossible 12');return
        print('Possible')
    else:print('Impossible');return
Box([3,3,4,4,4,4],[3,3,5,3,3,3])

UVA1588,有个陷阱要注意,注意有两种模拟的方向.直接模拟,先放入长一点的长条,然后从长的长条每一个点开始遍历.然后放入短的长条进行遍历,最后取最小值

def Kickdown(h1,h2):
    n1 = len(h1);n2 = len(h2)
    def geth(i):
        if(i>=n1):return 0
        else: return h1[i]
    def geth2(i):
        if(i>=n2):return 0
        else: return h2[i]
    for i in range(n1+1):
        if(geth(i)+h2[0]<=3):
            test = True
            for j in range(1,n2):
                if(geth(i+j)+h2[j]>3):test = False;break
            if(test):ans1 = max(n2+i,n1);break
    for i in range(n2+1):#遍历n2
        if(h1[0]+geth2(i)<=3):
            test = True
            for j in range(1,n1):
                if(h1[j]+geth2(i+j)>3):test = False;break
            if(test):ans2 = max(n1+i,n2);break
    print(min(ans1,ans2))

你可能感兴趣的:(算法竞赛入门经典第三章总结(2):后半部分习题解答)