AOJ 0033 Ball《挑战程序设计竞赛(第2版)》练习题答案 � 码农场


2.1 最基础的"穷竭搜索" 深度优先搜索
有一个形似央视大楼(Orz)的筒,从A口可以放球,放进去的球可通过挡板DE使其掉进B裤管或C裤管里,现有带1-10标号的球按给定顺序从A口放入,问是否有一种控制挡板的策略可以使B裤管和C裤管中的球从下往上标号递增。

输入:
第一行输入数据组数N。接下来N行为N组具体数据,每组数据中有10个整数,代表球的放入顺序。
输出:
对于每组数据,若策略存在,输出YES;若不存在,输出NO

Sample Input

2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
http://blog.csdn.net/synapse7/article/details/14454885
用二进制枚举即可。

int main(int argc, char *argv[])
{
    int n;
    cin >> n;
    while (n--)
    {
        int ball[10];
         
        for (int i = 0; i < 10; ++i)
        {
            cin >> ball[i];
        }
        bitset<10> direction;
        int all = 1024;
        while (all-- > 0)
        {
            direction = static_cast<bitset<10> >(all);
            bool perfect = true;
            int left = 0;
            int right = 0;
            for (int i = 0; i < 10; ++i)
            {
                if (direction[i])
                {
                    if (ball[i] > left)
                    {
                        left = ball[i];
                    }
                    else
                    {
                        perfect = false;
                        break;
                    }
                }
                else
                {
                    if (ball[i] > right)
                    {
                        right = ball[i];
                    }
                    else
                    {
                        perfect = false;
                        break;
                    }
                }
            }
            if (perfect)
            {
                break;
            }
        }
     
        if (all >= 0)
        {
            cout << "YES" << endl;
        }
        else
        {
            cout << "NO" << endl;
        }
    }
    return 0;
}
http://www.cnblogs.com/7hat/p/3590540.html
int a[10];

bool dfs(int k, int B, int C){
    if(k == 10) return true;
    if(a[k] > B){
        if(dfs(k + 1, a[k], C))
            return true;
    }
    if(a[k] > C){
        if(dfs(k + 1, B, a[k]))
            return true;
    }
    return false;
}

void solve(){
    if(dfs(0, 0, 0))
        printf("YES\n");
    else
        printf("NO\n");
}

int main(int argc, char const *argv[]){

    int t;
    scanf("%d", &t);
    while(t --){
        for(int i = 0; i < 10; i ++)
            scanf("%d", &a[i]);
        solve();
    }

    return 0;
}
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