USACO 1.1 – Broken Necklace | Jack Neus
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:
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.
假如你要在一些点打破项链,展开成一条直线，然后从一端开始收集同颜色的珠子直到你遇到一个不同的颜色珠子，在另一端做同样的事。(颜色可能与在这之前收集的不同) 确定应该在哪里打破项链来收集到最大多数的数目的子。举例来说，在图片 A 中的项链，可以收集到8个珠子,在珠子 9 和珠子 10 或珠子 24 和珠子 25 之间打断项链。 在一些项链中，包括白色的珠子如图片 B 所示。 当收集珠子的时候，一个被遇到的白色珠子可以被当做红色也可以被当做蓝色。表现项链的字符串将会包括三符号r，b和 w.写一个程序来确定从一条被供应的项链最大可以被收集珠子数目
X. O(N)
首先还是枚举断点，用bl,br,rl,rr来记录从第i个珠子开始，向左向右可以收集到的蓝色或红色珠子数目。对于边界问题，考虑最极端的情况是第n个珠子开始顺时针取n个，即全部取到，那么此时只要将项链再接上自己本身一段就可以实现了。同理，逆时针取的时候也是一样的。于是令s=s+s。
http://www.kkun.cc/articles/34
X. O(N^2) http://mangogao.com/read/46.html
http://code.antonio081014.com/2012/09/usaco-broken-necklace.html
Read full article from USACO 1.1 – Broken Necklace | Jack Neus
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:
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.
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)
11
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 ****** *****
假如你要在一些点打破项链,展开成一条直线，然后从一端开始收集同颜色的珠子直到你遇到一个不同的颜色珠子，在另一端做同样的事。(颜色可能与在这之前收集的不同) 确定应该在哪里打破项链来收集到最大多数的数目的子。举例来说，在图片 A 中的项链，可以收集到8个珠子,在珠子 9 和珠子 10 或珠子 24 和珠子 25 之间打断项链。 在一些项链中，包括白色的珠子如图片 B 所示。 当收集珠子的时候，一个被遇到的白色珠子可以被当做红色也可以被当做蓝色。表现项链的字符串将会包括三符号r，b和 w.写一个程序来确定从一条被供应的项链最大可以被收集珠子数目
X. O(N)
首先还是枚举断点，用bl,br,rl,rr来记录从第i个珠子开始，向左向右可以收集到的蓝色或红色珠子数目。对于边界问题，考虑最极端的情况是第n个珠子开始顺时针取n个，即全部取到，那么此时只要将项链再接上自己本身一段就可以实现了。同理，逆时针取的时候也是一样的。于是令s=s+s。
http://www.kkun.cc/articles/34
int main(){ ifstream fin("beads.in"); ofstream fout("beads.out"); int n; char tmp[400],s[800]; fin >> n >> tmp; //将项链复制一遍 for (int i=0;i<2*n;++i) s[i]=tmp[i%n]; int bl[800],br[800],rl[800],rr[800]; //向左 bl[0] = rl[0] = 0; for (int i=1;i<=2*n;++i){ if (s[i-1] == 'b'){ bl[i] = bl[i-1] + 1; rl[i] = 0; } else if (s[i-1] == 'r'){ bl[i] = 0; rl[i] = rl[i-1] + 1; } else { bl[i] = bl[i-1] + 1; rl[i] = rl[i-1] + 1; } } //向右 br[2*n] = rr[2*n] = 0; for (int i=2*n-1;i>=0;--i){ if (s[i] == 'b'){ br[i] = br[i+1] + 1; rr[i] = 0; } else if (s[i] == 'r'){ br[i] = 0; rr[i] = rr[i+1] + 1; } else { br[i] = br[i+1] + 1; rr[i] = rr[i+1] + 1; } } //求最大 int m(0); for (int i=0;i<2*n;++i) m = max(m, max(bl[i],rl[i]) + max(br[i],rr[i])); m = min(m,n); fout << m << endl; return 0;}X. O(N^2) http://mangogao.com/read/46.html
int n,p,ans(0);char beads[351];int next(int k, int r){ if (r == 1)//1为顺时针 if (k == n-1) return 0; else return k+1; if (r == -1)//-1为逆时针 if (k == 0) return n-1; else return k-1;}int cut(int k, int r){ int j; char color = 'w';//需要收集的珠子颜色，首先假定为w if (r == 1) p = k;//顺时针从当前珠子开始 if (r == -1) p = next(k,-1);//逆时针从前一个珠子开始 for (j=0;j<n;++j) { //取得需要收集的珠子颜色 if (color == 'w' && beads[p] != 'w') color = beads[p]; //已确定颜色、当前珠子不是白色、颜色错误时break if (color != 'w' && beads[p] != color && beads[p] != 'w') break; //取下一个珠子 p = next(p,r); } return j;}int main(){ ifstream fin("beads.in"); ofstream fout("beads.out"); fin >> n; for (int i = 0; i < n; ++i) fin >> beads[i]; int sum; for (int i = 0; i < n; ++i) { sum = cut(i,1) + cut(i,-1);//分别向两边取珠子 if (sum > ans) ans = sum;//判断最大 } if (ans > n) ans = n;//运气好会碰到两个方向能数到交叉 fout << ans << endl; return 0;}public void solve() throws Exception { BufferedReader br = new BufferedReader(new FileReader("beads.in")); PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter( "beads.out"))); int N = Integer.parseInt(br.readLine()); String strLine = br.readLine(); strLine += strLine; int max = 0; for (int i = 0; i < 2 * N; i++) { int m1 = 0; char a = 'w'; boolean first = true; for (int k = i; k < 2 * N; k++) { if (a == strLine.charAt(k) || strLine.charAt(k) == 'w') { m1++; if (strLine.charAt(k) != 'w') { first = false; a = strLine.charAt(k); } } else if (strLine.charAt(k) != 'w') { if (first) { first = false; m1++; a = strLine.charAt(k); } else break; } } a = 'w'; first = true; int m2 = 0; for (int k = i - 1; k >= 0; k--) { if (a == strLine.charAt(k) || strLine.charAt(k) == 'w') { m2++; if (strLine.charAt(k) != 'w') { first = false; a = strLine.charAt(k); } } else if (strLine.charAt(k) != 'w') { if (first) { first = false; m2++; a = strLine.charAt(k); } else break; } } if (m1 + m2 > max) { max = m1 + m2; // System.out.println(i + ": " + strLine.charAt(i) + ", " + // max); // System.out.println(m1 + ", " + m2); } if (m1 + m2 >= N) { max = N; break; } } // System.out.println(max); out.write("" + max + "\n"); out.close(); }
Read full article from USACO 1.1 – Broken Necklace | Jack Neus
