Broken Necklace

最新推荐文章于 2024-10-20 21:10:12 发布

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最新推荐文章于 2024-10-20 21:10:12 发布

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分类专栏： USACO 文章标签： string output include file input each

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针对一个给定的红、蓝、白色珠子项链，通过算法确定最佳断裂点以收集最多数量的相同颜色珠子，考虑白色珠子可以当作红色或蓝色的情况。

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Problem:

Broken Necklace

You have a necklace of N red, white, or blue beads (3<=N<=350) some of which are red, others blue, and others white, arranged at random. Here are two examples for n=29:

                1 2                               1 2
            r b b r                           b r r b
          r         b                       b         b
         r           r                     b           r
        r             r                   w             r
       b               r                 w               w
      b                 b               r                 r
      b                 b               b                 b
      b                 b               r                 b
       r               r                 b               r
        b             r                   r             r
         b           r                     r           r
           r       r                         r       b
             r b r                             r r w
            Figure A                         Figure B
                        r red bead
                        b blue bead
                        w white bead

The beads considered first and second in the text that follows have been marked in the picture.

The configuration in Figure A may be represented as a string of b's and r's, where b represents a blue bead and r represents a red one, as follows: brbrrrbbbrrrrrbrrbbrbbbbrrrrb .

Suppose you are to break the necklace at some point, lay it out straight, and then collect beads of the same color from one end until you reach a bead of a different color, and do the same for the other end (which might not be of the same color as the beads collected before this).

Determine the point where the necklace should be broken so that the most number of beads can be collected.

Example

For example, for the necklace in Figure A, 8 beads can be collected, with the breaking point either between bead 9 and bead 10 or else between bead 24 and bead 25.

In some necklaces, white beads had been included as shown in Figure B above. When collecting beads, a white bead that is encountered may be treated as either red or blue and then painted with the desired color. The string that represents this configuration will include the three symbols r, b and w.

Write a program to determine the largest number of beads that can be collected from a supplied necklace.

PROGRAM NAME: beads

INPUT FORMAT

Line 1:	N, the number of beads
Line 2:	a string of N characters, each of which is r, b, or w

SAMPLE INPUT (file beads.in)

29
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb

OUTPUT FORMAT

A single line containing the maximum of number of beads that can be collected from the supplied necklace.

SAMPLE OUTPUT (file beads.out)

OUTPUT EXPLANATION

Consider two copies of the beads (kind of like being able to runaround the ends). The string of 11 is marked.

wwwbbrwrbrbrrbrbrwrwwrbwrwrrb wwwbbrwrbrbrrbrbrwrwwrbwrwrrb
                       ****** *****

My Answer:

/* ID : Iby071 PROG : beads LANG : C++ */ # include<iostream> # include<fstream> using namespace std; int main() { ifstream fin("beads.in"); ofstream fout("beads.out"); int N,rTotal=0,bTotal=0,wTotal=0,max=0; int i,j; char ch; fin>>N; int *num=new int[3*N]; for(i=0;i<N;++i) { fin>>ch; switch(ch) { case 'r':num[i]=num[N+i]=num[2*N+i]=0;break; case 'b':num[i]=num[N+i]=num[2*N+i]=1;break; case 'w':num[i]=num[N+i]=num[2*N+i]=2;break; default: break; } } for( i=N;i<2*N;++i) { int tmp=i; //记录使便于还原i if( num[i]==0 ) { for(j=i; num[j]!=1&&j<3*N; ++j) ++rTotal; for( ;num[j]!=0&&j<3*N;++j) ++bTotal; --i; for( ;num[i]!=1&&i>=0;--i) ++rTotal; if(rTotal+bTotal>max) max=rTotal+bTotal; //解决可能出现的全是r和w或全是b和w的特殊情况 } else if( num[i]==1 ) { for(j=i; num[j]!=0&&j<3*N; ++j) ++bTotal; for( ;num[j]!=1&&j<3*N;++j) ++rTotal; --i; for( ;num[i]!=0&&i>=0;--i) ++bTotal; if(rTotal+bTotal>max) max=rTotal+bTotal; } else ++wTotal; i=tmp; rTotal=bTotal=0; //开始新一次断链和计数前，给i，rTotal和bTotal复原 } if(max>N||wTotal==N) max=N; //解决可能出现的全是w的特殊情况 fout<<max<<endl; return 0;

My Gain:

写这个程序时，第一次遇到因为运行时间过长而通不过。

我当时写了一个循环顺序表来存放这些珠子。

后来在网络上看到有人说，可以不用做成那样，直接三个一模一样的数组，第一个的后一半+第二个数组+第三个的前一半，就足够了。

有心计啊^_^ORZ...