HDU 5532 Almost Sorted Array(狀態壓縮DP)

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2526    Accepted Submission(s): 627

Problem Description

We are all familiar with sorting algorithms: quick sort, merge sort, heap sort, insertion sort, selection sort, bubble sort, etc. But sometimes it is an overkill to use these algorithms for an almost sorted array.

We say an array is sorted if its elements are in non-decreasing order or non-increasing order. We say an array is almost sorted if we can remove exactly one element from it, and the remaining array is sorted. Now you are given an array a1,a2,…,an, is it almost sorted?

 

Input

The first line contains an integer T indicating the total number of test cases. Each test case starts with an integer n in one line, then one line with n integers a1,a2,…,an.

1≤T≤2000
2≤n≤105
1≤ai≤105
There are at most 20 test cases with n>1000.

 

Output

For each test case, please output "`YES`" if it is almost sorted. Otherwise, output "`NO`" (both without quotes).

 

Sample Input

3 3 2 1 7 3 3 2 1 5 3 1 4 1 5

 

Sample Output

YES YES NO

 

Source

2015ACM/ICPC亞洲區長春站-重現賽（感謝東北師大）

 

Recommend

hujie

我們分析每一行種植玉米的狀態其實之和前一行的種植狀態有關係，和它前面的其餘行沒關係，這個滿足無後效性，
題目讓我們求的是方案總數，其實就是求最後一行的所有狀態的方案數加和。
令dp[i][state]表示第i行狀態爲state的方案數，那麼dp[i][state] += dp[i-1][pre_state],這裏pre_state和state
必須滿足條件，1：不相鄰 2：pre_state和state是可以存在的，因爲
有的地不能種植玉米，相應位不能爲1，該轉移方程可以理解爲前一行的每一個pre_state都爲第i行達到狀態state貢
獻了dp[i-1][pre_state]個方案數。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=13;
int dp[N][1<<N];
int n,m,a[N][N];
const int mod=100000000;
int vst[1<<N],num;
void getvalidstate()
{
    num=0;
    for(int i=0;i<(1<<n);i++){
        if((i&(i<<1)) == 0)
            vst[num++]=i;
    }
}
bool valid(int state,int row)
{
    int flag=0;
    for(int i=0;i<n;i++)
    {
        if(!a[row][i+1]){
            if(state&(1<<i))
            {
                flag=1;
            }
        }
    }
    if(flag) return false;
    else return true;
}
int main()
{
    while(~scanf("%d %d",&m,&n))
    {
        memset(a,0,sizeof(a));
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
              scanf("%d",&a[i][j]);
        dp[0][0]=1;
        getvalidstate();
        for(int i=1;i<=m;i++) /// 枚舉每一行
        {
            for(int j=0;j<num;j++) /// 枚舉每一個滿足條件的狀態
            {
                 if(valid(vst[j],i)){
                    for(int k=0;k<num;k++){
                        if(valid(vst[k],i-1) && !(vst[j]&vst[k])){
                            dp[i][vst[j]]+=dp[i-1][vst[k]];
                            dp[i][vst[j]]%=mod;
                        }
                    }
                 }
            }
        }
        int ans=0;
        for(int i=0;i<(1<<n);i++)
        {
            ans=(ans+dp[m][i])%mod;
        }
        printf("%d\n",ans);
    }
    return 0;
}