pku 3087 Shuffle'm Up 说的是bfs，其实就是个模拟

http://poj.org/problem?id=3087

刚看到题目的时候给我整蒙了，BFS怎么做啊？这不就是模拟一下洗牌的过程，如果遇到目标字符串就输出步数，如果回到原始串，就说明到不了目标串。

View Code

#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

using namespace std;



char aim[207];//目标串

int n;



struct node

{

    char s1[107];

    char s2[107];

    int len;

};

bool isok(char *s,int start)//检查串

{

    for (int i = start, j = 0; j < n; ++j,++i)

    {

        if (aim[i] != s[j]) return false;

    }

    return true;

}

int bfs(char *s1,char *s2)

{

    node s;

    queue<node>q;

    strcpy(s.s1,s1);

    strcpy(s.s2,s2);

    s.len = 0;

    q.push(s);

    int sum = -1;

    bool flag = false;//拥有标记不是初始串

    //int test  = 10;

    while (!q.empty())

    {

        node cur = q.front(); q.pop();

        if (isok(cur.s1,0) && isok(cur.s2,n))//找到目标串

        {

            sum = cur.len;

            break;

        }

        if (flag && !strcmp(cur.s1,s1) && !strcmp(cur.s2,s2))  break;//回到原始串

        flag = true;

        char str[207];

        int i = 0;

        int l = 0;

        node tt;

        for (i = 0; i < n; ++i)

        {

            str[l++] = cur.s2[i];

            str[l++] = cur.s1[i];

        }

        str[l] = '\0';

        //printf(">>>>>%s\n",str);

        for (i = 0; i < n; ++i)  tt.s1[i] = str[i];

        tt.s1[n] = '\0';

        for (i = n; i < 2*n; ++i) tt.s2[i - n] = str[i];

        tt.s2[n] = '\0';

        //printf(">>>>>%s %s %s\n",str,tt.s1,tt.s2);

        tt.len = cur.len + 1;

        q.push(tt);

    }

    return sum;

}

int main()

{

    char s1[107],s2[107];

    int t,cas = 1;

    scanf("%d",&t);

    while (t--)

    {

        scanf("%d",&n);

        scanf("%s%s",s1,s2);

        scanf("%s",aim);

        int flag = bfs(s1,s2);

        printf("%d %d\n",cas++,flag);

    }

    return 0;

}