POJ 2893 M × N Puzzle-- eight digital solvability conditions

Meaning of the questions: Given M * N digital map, asked if he could move to the final state

analysis

Solving a condition seen eight digital solvability condition

It is worth mentioning that this question seeking to reverse the card Fenwick tree, with only merge sort.

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

const int maxn = 1e6 + 10;
int n,m,pos,x,ans,zero;
int seq[maxn],tmp[maxn];

void msort(int l,int r)  //对seq进行排序
{
    if(l==r) return ;
    int mid=(l+r)>>1;
    msort(l,mid);
    msort(mid+1,r);
    int i=l,j=mid+1,k=l-1;
    while(i<=mid&&j<=r)
    {
        if(seq[i]<=seq[j])
            tmp[++k]=seq[i],i++;
        else tmp[++k]=seq[j],j++,ans+=mid-i+1;  //ans就是逆序对个数
    }
    while(i<=mid)
        tmp[++k]=seq[i],i++;
    while(j<=r)
        tmp[++k]=seq[j],j++;
    for(int qwq=l;qwq<=r;qwq++)
        seq[qwq]=tmp[qwq];
}

int main()
{
    while(scanf("%d%d",&n,&m)==2 && m)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
            {
                scanf("%d",&x);
                if(x) seq[++pos]=x;
                else zero=n-i;
            }
        msort(1,pos);
        if(m&1)  zero=0;
        if((ans+zero)%2==0)  printf("YES\n");
        else printf("NO\n");
        ans=0;pos=0;zero=0;
    }
    return 0;
}

 

 

Reference Links: https://www.cnblogs.com/nopartyfoucaodong/p/9673434.html