hihocoder 1233 Boxes

题意：给出n个不同的数，每次可以把一个较小的那个数移到相邻的较大的那个数的上面，问最少需要移动多少次使得这n个数是升序的。

做法：从升序状态bfs出到其它状态的最小步数即可，状态s:从左到右，第i个数在第j个位置，例如

序列312表示成231，因为1在第2个位置，2在第3个位置，3在第1个位置。

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
int *dp[8];
int a[8],fac[8];
void bfs(int n)
{
int x=0;
for(int i=1;i<=n;i++)
x=x*10+i;
queue<int>qq;
qq.push(x);
dp
[x]=0;
while(qq.size())
{
int from=qq.front();
qq.pop();
memset(a,0,sizeof(a));
x=from;
for(int i=n;i>0;i--,x/=10)
a[x%10]=i;
for(int i=1;i<=n;i++)
if(a[i]!=0)
{
if(i>1&&(a[i]<a[i-1]||a[i-1]==0))
{
int to=from-fac[n-a[i]];
if(dp
[to]==-1)
{
qq.push(to);
dp
[to]=dp
[from]+1;
}
}
if(i<n&&(a[i]<a[i+1]||a[i+1]==0))
{
int to=from+fac[n-a[i]];
if(dp
[to]==-1)
{
qq.push(to);
dp
[to]=dp
[from]+1;
}
}
}
}
}
int main()
{
fac[0]=1;
for(int i=1;i<=7;i++)
fac[i]=fac[i-1]*10;
int x=0;
for(int i=1;i<=7;i++)
{
x=x*10+8;
dp[i]=new int[x+1];
memset(dp[i],-1,sizeof(int)*x);
bfs(i);
}
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",a+i);
x=0;
for(int i=1;i<=n;i++)
{
int sum=0;
for(int j=1;j<=n;j++)
if(a[i]>=a[j])
sum++;
x+=i*fac[n-sum];
}
printf("%d\n",dp
[x]);
}
}

时间限制:1000ms
单点时限:1000ms
内存限制:256MB

描述

There is a strange storehouse in PKU. In this storehouse there are n slots for boxes, forming a line. In each slot you can pile up any amount of boxes. The limitation is that you can only pile a smaller one above
a bigger one, in order to keep balance. The slots are numbered from 1 to n. The leftmost one is slot 1.

At first there is exactly one box in each slot. The volume of the box in slot i is vi. As a virgo, you
decide to sort these boxes by moving some of them. In each move you can choose a slot and move the top box in this slot to an adjacent slot (of course you can only put it on the top). You should ensure that the limitation mentioned above is still satisfied
after this move. After the sort operation, there should be exactly one box in each slot, and for each pair of adjacent slots, the box in the left one should be smaller than the box in the right one.
Your task is to calculate the minimum number of moves you need to sort the boxes.

输入

In the first line there’s an integer T(T≤6000), indicating the number of test cases. The following 2T lines describe the test cases.
In each test case, the first line contains an integer n, indicating the number of slots. The second line contains n integers v1,v2…vn,
indicating the volume of the boxes. It is guaranteed that all vi in a test case are different.
Please note that n<8,0≤vi≤104

输出

For each test case, print a line containing one integer indicating the answer. If it’s impossible to sort the boxes, print -1.

样例输入

4
3
2 1 3
2
7 8
2
10000 1000
3
97 96 95

样例输出

4
0
-1
20