做这题做了一个下午,都快奔溃了。说下要点:
1、要注意'w'这种特殊情况,到某位置时,注意左边、右边是'w'的情况。
2、得把整个项链看出循环队列,所以指针移动时注意mod。
3、我用的还是暴力法,O(n^2),看解题报告有O(n)的算法,用的是DP,还有待研究。
源程序
#define _CRT_SECURE_NO_WARNINGS #include <iostream> #include <cstdio> #include <string> #include <algorithm> using namespace std; struct Bead { char color; int num; }; int len; int inc(int i) { return (i + 1) % len; } int dec(int i) { return (len + i - 1) % len; } int main(int argc, char **argv) { freopen("beads.in", "r", stdin); freopen("beads.out", "w", stdout); cin >> len; string str; cin >> str; int max = 0; for (size_t i = 0; i < str.length(); ++i) { int tmp = 0; int L = dec(i); int R = i; if (str[L] == str[R]) { //continue; } if (str[R] == 'w') { int j; ++tmp; // str[R]总是计入 // 消除重复的'w' for (j = inc(R); j != L && str[j] == 'w'; j = inc(j)) { ++tmp; } if (str[j] != 'w') { int index = j; ++tmp; // 连续的'w'后的字符计入 for (j = inc(j); j != L && (str[j] == str[index] || str[j] == 'w'); j = inc(j)) { ++tmp; } } } else { ++tmp; for (int j = inc(R); j != L && (str[j] == str[R] || str[j] == 'w'); j = inc(j)) { ++tmp; } } if (str[L] == 'w') { int j; ++tmp; // str[L]总是计入 // 消除重复的'w' for (j = dec(L); j != R && str[j] == 'w'; j = dec(j)) { ++tmp; } if (str[j] != 'w') { int index = j; ++tmp; // 连续的'w'前的字符计入 for (j = dec(j); j != R && (str[j] == str[index] || str[j] == 'w'); j = dec(j)) { ++tmp; } } } else { ++tmp; for (int j = dec(L); j != R && (str[j] == str[L] || str[j] == 'w'); j = dec(j)) { ++tmp; } } if (tmp > max) { max = tmp; } } cout << min(max, len) << endl; return 0; }
附:原题链接:http://cerberus.delos.com:790/usacoprob2?a=7jIrrj5a9HY&S=beads
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)
11
OUTPUT EXPLANATION
Consider two copies of the beads (kind of like being able to runaround the ends). The string of 11 is marked.original 'split' v wwwbbrwrbrbrrbrbrwrwwrbwrwrrb|wwwbbrwrbrbrrbrbrwrwwrbwrwrrb ******|***** rrrrrb|bbbbb <-- assignments 5 x r 6 x b <-- 11 total