Catch That Cow（HDU 2717）

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Catch That Cow

题目就不摘了-。-

题目链接

这道题一般用广搜写，可以用循环队列或队列模板来解决数组开得不够大的问题。

测试时试试 0 10000这个数据，其结果为22，一定要保证 0 10000 不会栈溢出，否则出现running time error

本题关键是在入队之前就进行判断，这样就会减少入队的次数。

1，循环队列(c，c++)

#include<cstdio>#include<cmath>#include<iostream>#include<algorithm>#include<cstring>#include<string.h>#include<stdlib.h>using namespace std;int main(){    int n, k;    struct node    {        int s;        int num;    }que[200010];//队列    int book[200010];//用来标记    while (scanf("%d%d", &n, &k) == 2)    {        if (n >= k)            printf("%d\n", n - k);//负的直接输出        else        {            int min = k - n;            memset(book, 0, sizeof(book));            memset(que, 0, sizeof(que));            int head, tail;            head = 1;            tail = 1;            que[tail].num = n;            que[tail].s = 0;            tail++;            while (head < tail)            {                book[que[head].num] = 1;                que[tail].num = que[head].num + 1;                if (que[tail].num == k)                    break;                if (book[que[tail].num] == 0                       && que[tail].num >= 0                       && que[tail].num <= 100000)//入队判断条件有三个                {                    book[que[tail].num] = 1;                    que[tail++].s = que[head].s + 1;                }                if (tail >= 200000)                    tail = 1;                //接下来两个方向相当于复制-粘贴                que[tail].num = que[head].num * 2;                if (que[tail].num == k)                    break;                if (book[que[tail].num] == 0                       && que[tail].num >= 0                       && que[tail].num <= 100000)                {                    book[que[tail].num] = 1;                    que[tail++].s = que[head].s + 1;                }                if (tail >= 200000)                    tail = 1;                que[tail].s = que[head].s + 1;                que[tail].num = que[head].num - 1;                if (que[tail].num == k)                    break;                if (book[que[tail].num] == 0                       && que[tail].num >= 0                       && que[tail].num <= 100000)                {                    book[que[tail].num] = 1;                    que[tail++].s = que[head].s + 1;                }                if (tail >= 200000)                    tail = 1;                head++;                 if (head >= 200000)                    head = 1;            }            printf("%d\n", que[head].s+1);        }    }    return 0;}

2，队列模板(c++)

#include<cstdio>#include<cmath>#include<iostream>#include<algorithm>#include<cstring>#include<string.h>#include<stdlib.h>#include<queue>using namespace std;int cow[100010];int n, k;void bfs(){    cow[n] = 0;    queue<int> Q;    int x, y;    Q.push(n);//入队    while (!Q.empty())//head == tail ??    {        x = Q.front();//返回队首        Q.pop();//出队        if (x == k)            return;        y = x + 1;        if (y >= 0 && y <= 100000 && cow[y] == 0)        {            cow[y] = cow[x] + 1;            Q.push(y);        }        y = x - 1;        if (y >= 0 && y <= 100000 && cow[y] == 0)        {            cow[y] = cow[x] + 1;            Q.push(y);        }        y = x * 2;        if (y >= 0 && y <= 100000 && cow[y] == 0)        {            cow[y] = cow[x] + 1;            Q.push(y);        }    }}int main(){    while (cin >> n >> k)    {        memset(cow, 0, sizeof(cow));        bfs();        cout << cow[k] << endl;    }    return 0;}

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