Broken Necklace(USACO官方)

复杂度为O(n²):

#include <stdio.h> #include <string.h> #include <assert.h>#define MAXN 400char necklace[MAXN]; int len;/* * Return n mod m. The C % operator is not enough because* its behavior is undefined on negative numbers.*/ //这种处理转一圈回到头的情况很值得学习 int mod(int n, int m) {while(n < 0)n += m;return n%m; }/** Calculate number of beads gotten by breaking* before character p and going in direction dir,* which is 1 for forward and -1 for backward.*/ //dir为方向 int nbreak(int p, int dir) {char color;int i, n;//用color来保存颜色color = 'w';/* Start at p if going forward, bead before if going backward */if(dir > 0)i = p;elsei = mod(p-1, len);/* We use "n<len" to cut off loops that go around the whole necklace */for(n=0; n<len; n++, i=mod(i+dir, len)) {/* record which color we're going to collect *///处理收集珠子时颜色的变换,处理得比我那函数好多了if(color == 'w' && necklace[i] != 'w')color = necklace[i];/* * If we've chosen a color and see a bead* not white and not that color, stop */if(color != 'w' && necklace[i] != 'w' && necklace[i] != color)break;}return n; }void main(void) {FILE *fin, *fout;int i, n, m;fin = fopen("beads.in", "r");fout = fopen("beads.out", "w");assert(fin != NULL && fout != NULL);fscanf(fin, "%d %s", &len, necklace);assert(strlen(necklace) == len);m = 0;for(i=0; i<len; i++) {n = nbreak(i, 1) + nbreak(i, -1);if(n > m)m = n;}/** If the whole necklace can be gotten with a good* break, we'll sometimes count beads more than * once. this can only happen when the whole necklace* can be taken, when beads that can be grabbed from* the right of the break can also be grabbed from the left.*///考虑到转一圈的情况if(m > len)m = len;fprintf(fout, "%d/n", m);exit (0); }

官方答案2:复杂度O(n)

#include <stdio.h> #include <string.h> #include <algorithm>using namespace std;FILE *in,*out;int main () {in = fopen("beads.in", "r");out = fopen ("beads.out", "w");int n;char tmp[400], s[800];fscanf(in, "%d %s", &n, tmp);strcpy(s, tmp);//也是珠子加上它自己的一个副本strcat(s, tmp);//用0列来保存红珠子的数目,用1列来保存蓝珠子的数目int left[800][2], right[800][2];left[0][0] = left[0][1] = 0;//其中一个方向for (int i=1; i<= 2 * n; i++){if (s[i - 1] == 'r'){left[i][0] = left[i - 1][0] + 1;left[i][1] = 0;} else if (s[i - 1] == 'b'){left[i][1] = left[i - 1][1] + 1;left[i][0] = 0;} else {left[i][0] = left[i - 1][0] + 1;left[i][1] = left[i - 1][1] + 1;}}//另一个方向right[2 * n][0] = right[2 * n][1] = 0;for (int i=2 * n - 1; i >= 0; i--){if (s[i] == 'r'){right[i][0] = right[i + 1][0] + 1;right[i][1] = 0;} else if (s[i] == 'b'){right[i][1] = right[i + 1][1] + 1;right[i][0] = 0;} else {right[i][0] = right[i + 1][0] + 1;right[i][1] = right[i + 1][1] + 1;}}int m = 0;//这里很牛X,注意向左向右都是i位置的for (int i=0; i<2 * n; i++)m = max(m, max(left[i][0], left[i][1]) + max(right[i][0], right[i][1]));//转一圈的情况m = min(m, n);fprintf(out, "%d/n", m);fclose(in); fclose(out);return 0; }

Dynamic Programming is good method for solving this problem in O(N). If we consider two copies of the string we easy transform cyclic configuration of the necklace to linear. Now we can compute for each breaking point how many beads of the same color can be collected on the left and on the right from the breaking point. I show how we can compute it only for the left side. For right side it is analogical. Let r[p] and b[p] be the number of red / blue beads that can be collected, when necklace is broken in point p. If we know this and color of next bead (c) we can compute r[p+1] and b[p+1].

 r[0] = p[0] = 0If c = 'r' then r[p+1] = r[p] + 1 and b[p+1] = 0because the length of the blue beads is 0.if c = 'b' then b[p+1] = b[p] + 1 and r[p+1] = 0if c = 'w' then both length of the red and length of blue beadscan be longer.
so r[p+1] = r[p]+1 and b[p+1] = b[p] + 1.

The number of beads that can be collected in breaking point p is then max(left[r[p]], left[b[p]]) + max(right[r[p]], right[b[p]]). And the maximum from this value is answer for the problem.

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