USACO 1.1.4 Broken Necklace(beads)

最新推荐文章于 2020-07-19 16:42:58 发布

Gavin_dinggengjia

最新推荐文章于 2020-07-19 16:42:58 发布

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分类专栏： USACO 文章标签： null file

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本文介绍了一种解决特定串链珠子问题的方法，通过将环形链转化为直线型并利用双指针技巧来计算最大连续相同颜色珠子的数量。代码采用C语言实现，并考虑了边界条件以确保准确性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1、串联两条链解开环形

2、暴力搜索

3、注意边界条件

rrr

易出错！

/* ID: gengjia1 LANG: C TASK: beads */ #include <stdio.h> #include <stdlib.h> #include <string.h> #include <assert.h> int main(void) { FILE *fin = fopen ("beads.in", "r"); FILE *fout = fopen ("beads.out", "w"); int N, n = 0; char *s = NULL, *st = NULL; int fh = 0, sh = 0; //first half, second half char fc, lc, cc; //first half char, last char, current char int i, j; fscanf (fin, "%d", &N); assert(N >= 3); assert(N <= 350); s = (char *)malloc(2 * N + 1); st = (char *)malloc(N+1); if( s == NULL || st == NULL) exit(1); fscanf (fin, "%s", st); assert(strlen(st) == N); s = strcpy(s, st); s = strcat(s, st); for(i=0; i < N; i++) { fh = 0; sh = 0; j = 0; fc = lc = s[i]; fh += 1; j = i + 1; cc = s[j]; while( !(sh > 0 && cc == fc) ) //前半段字符第二次出现相等 { if(cc == lc) { if(sh == 0) { if(fh >= N) { break; } fh += 1; } else { sh +=1; } } else { if(fc == lc) { if(fc == 'w') { fh += 1; fc = cc; } else if(cc =='w') { fh += 1; } else { sh += 1; } } else { if(fc == cc) { fh += 1; } else { sh += 1; } } lc = cc; } j += 1; cc = s[j]; } if (n < fh + sh) { n = fh + sh; } if (n == N) { break; //注意边界条件 } } fprintf (fout, "%d/n", n); free(s); fclose(fin); fclose(fout); exit(0); }