Wiki OI 1099 字串变换

题目链接：http://wikioi.com/problem/1099/

算法与思路：双向广搜

所谓双向搜索指的是搜索沿两个方向同时进行：

正向搜索：从初始结点向目标结点方向搜索；

逆向搜索：从目标结点向初始结点方向搜索；

当两个方向的搜索生成同一子结点时终止此搜索过程。

详见注释。

#include<stdio.h>#include<string.h>struct node{    char s[30];    int dep;  //变换次数 } list1[5010], list2[5010];char a[7][30], b[7][30];int n;void BFS(){    int head1, tail1, head2, tail2, k;    head1 = tail1 = head2 = tail2 = 1;    while(head1 <= tail1 && head2 <= tail2)    {        if(list1[head1].dep + list2[head2].dep > 10)         {            printf("NO ANSWER!\n");            return ;        }        for(int i = 0;i < strlen(list1[head1].s); i++)            for(int j = 1; j <= n; j++)                if(strncmp(list1[head1].s + i, a[j], strlen(a[j])) == 0) //寻找当前可变换的规则                 {                  tail1++; //移动尾指针，存储变换后的字符串，以下三个for循环为变换过程                   for(k = 0; k < i; k++)       list1[tail1].s[k] = list1[head1].s[k];                  for(int l = 0; l < strlen(b[j]); l++, k++)       list1[tail1].s[k] = b[j][l];                  for(int l = i + strlen(a[j]); l <= strlen(list1[head1].s); l++, k++)                     list1[tail1].s[k] = list1[head1].s[l];                  list1[tail1].s[k] = '\0'; //为变换结束后的字符串加结束符                   list1[tail1].dep = list1[head1].dep+1;                  for (k = 1; k <= tail1; k++)                    if (strcmp(list1[tail1].s, list2[k].s) == 0)//判断当前状态是否与逆向搜索交汇                     {                       printf("%d\n", list1[tail1].dep + list2[k].dep);                       return ;                     }                }        for (int i = 0; i < strlen(list2[head2].s); i++) //逆向搜索同上             for (int j = 1; j <= n; j++)                if(strncmp(list2[head2].s + i, b[j], strlen(b[j])) == 0)                {                  tail2++;                  for(k = 0; k < i; k++)       list2[tail2].s[k] = list2[head2].s[k];                  for(int l = 0; l < strlen(a[j]); l++, k++)       list2[tail2].s[k] = a[j][l];                  for(int l = i + strlen(b[j]); l <= strlen(list2[head2].s); l++, k++)                    list2[tail2].s[k] = list2[head2].s[l];                  list2[tail2].s[k] = '\0';                  list2[tail2].dep = list2[head2].dep + 1;                  for (k = 1;k <= tail1; k++)                    if (strcmp(list1[k].s, list2[tail2].s) == 0)                    {                       printf("%d\n",list1[k].dep + list2[tail2].dep);                       return ;                    }                }        head1++; head2++;    }    printf("NO ANSWER!\n");}int main(){    scanf("%s%s",list1[1].s, list2[1].s);    n = 1;    while (scanf("%s%s",a[n],b[n]) != EOF)     n++;n--;    list1[1].dep = list2[1].dep = 0;    BFS();    return 0;}