usaco1.1.4 Broken Necklace 题解

【算法】模拟　　【难度】★☆☆☆☆

仍然是模拟，但要注意这个项链是环状的。有一种比较先进的办法是把这个项链载入两次，这样就可以从头到尾遍历了。注意白色的珠子不能作为起点。
我的程序没有用上面的方法，直接遍历到最后在返回开头。
View Code 
 1 /*
 2 ID: wsc5001
 3 LANG: C
 4 TASK: beads
 5 */
 6 #include<stdio.h>
 7 #include <stdlib.h>
 8 char ch[351];
 9 int n;
10 int zhaoqian(int st)
11 {
12     char goon;
13     int i=0,timess=0;
14     while(st+i<n && (ch[st+i]=='w') )
15         {i++;}
16     while(st+i>=n && (ch[st+i-n]=='w'))
17         {i++;}
18     goon = ch[st+i];
19     i=0;
20     while(st+i<n && (ch[st+i]==goon||ch[st+i]=='w'))
21         {timess++; i++;}
22     while(st+i>=n && (ch[st+i-n]==goon||ch[st+i-n]=='w'))
23         {timess++; i++;}
24     return timess;
25 }
26 int zhaohou(int st)
27 {
28     char goon;
29     int i=0,timess=0;
30     while(st-i>=0 && (ch[st-i]=='w'))
31     {i++;}
32     while(st-i<0 && (ch[st-i+n]=='w'))
33     {i++;}
34     goon = ch[st-i];
35     i=0;
36     while( st-i>=0 && (ch[st-i]==goon||ch[st-i]=='w') )
37         {timess++; i++;}
38     while(st-i<0 && (ch[st-i+n]==goon||ch[st-i+n]=='w'))
39         {timess++; i++;}
40     return timess;
41 }
42 int main()
43 {
44     FILE *fin,*fout;
45     fin=fopen("beads.in","r");
46     fout=fopen("beads.out","w");
47     int nfront=0,nback=0,max=0;
48     int i,j,flag;
49     //读入 
50     fscanf(fin,"%d%s",&n,ch);
51     flag=0;
52     for (i=0;i<n;i++)
53         if(ch[i]!=ch[0]&&ch[i]!='w')
54         {flag=1;break;}
55     if (flag==0)
56     {fprintf(fout,"%d\n",n);}
57     else
58     {
59         for (i=0;i<n;i++)
60         {
61             nfront=zhaoqian(i+1);
62             nback=zhaohou(i);
63             if (nfront+nback>max)
64                 max=nfront+nback;
65             if (max>n)
66                 max=n;
67         }
68     fprintf(fout,"%d\n",max);
69     }
70     fclose(fin);
71     fclose(fout);
72     //system("pause");
73 }


posted @ 2012-01-10 21:47  wsc500  阅读(291)  评论(0)  编辑  收藏  举报