noip模拟赛 PA

分析：很显然这是一道搜索题，可能是由于我的搜索打的太不美观了，这道题又WA又T......如果对每一个询问都做一次bfs是肯定会T的，注意到前70%的数据范围，N的值都相等，我们可以把给定N的所有情况给算出来，然后O（1）查询.从终点状态往起点状态BFS就可以了.

当最终状态很少，起始状态很多的时候，可以考虑倒着做.

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int T, jian[], stu, shu[], cnt, top[], d[], cur[], vis[];
int n, a[], b[];
queue <int> q;

bool cmp(int x, int y)
{
    return a[x] < a[y];
}

void bfs()
{
    while (!q.empty())
    {
        int zhuangtai = q.front();
        q.pop();
        memset(top, , sizeof(top));
        memset(shu, , sizeof(shu));
        cnt = ;
        int tt = zhuangtai;
        while (tt)
        {
            shu[++cnt] = tt % ;
            tt /= ;
        }
        reverse(shu + , shu +  + cnt);
        for (int i = cnt; i >= ; i--)
            top[shu[i]] = i;
        for (int i = ; i <= cnt; i++)
            if (i == top[shu[i]])
            {
            int temp = shu[i];
            if (temp !=  && (i < top[temp - ] || !top[temp - ]))
            {
                int newstu = zhuangtai - jian[cnt - i];
                if (!vis[newstu])
                {
                    q.push(newstu);
                    vis[newstu] = ;
                    d[newstu] = d[zhuangtai] + ;
                }
            }

            if (temp != cnt && (i < top[temp + ] || !top[temp + ]))
            {
                int newstu = zhuangtai + jian[cnt - i];
                if (!vis[newstu])
                {
                    q.push(newstu);
                    vis[newstu] = ;
                    d[newstu] = d[zhuangtai] + ;
                }
            }
            }
    }
}

int main()
{
    scanf("%d", &T);
    jian[] = ;
    stu = ;
    for (int i = ; i <= ; i++)
    {
        jian[i] = jian[i - ] * ;
        stu = stu *  + i;
        q.push(stu);
        vis[stu] = ;
    }
    bfs();
    //printf("flag\n");
    while (T--)
    {
        scanf("%d", &n);
        for (int i = ; i <= n; i++)
        {
            scanf("%d", &a[i]);
            b[i] = i;
        }
        sort(b + , b +  + n, cmp);
        int stu = ;
        for (int i = ; i <= n; i++)
            stu = stu *  + b[i];
        if (!vis[stu])
            printf("-1\n");
        else
            printf("%d\n", d[stu]);
    }

    return ;
}