M × N Puzzle

来自逆序对的强大力量
 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<algorithm>
 4 #include<cstring>
 5 #define lowbit(x)(x&-x)
 6 using namespace std;
 7 int c[1000000],a[1000][1000],maxn,n,m,d[1000000];
 8 void update(int k,int x){for(int i=k;i<=maxn;i+=lowbit(i))c[i]+=x;}
 9 int getsum(int x)
10 {
11     int ans=0;
12     for(int i=x;i;i-=lowbit(i))ans+=c[i];
13     return ans;
14 }
15 int main()
16 {
17     while(1)
18     {
19         
20         int cnt=0,num=0;
21         scanf("%d%d",&m,&n);
22         if(m==0&&n==0) break;
23         maxn=m*n;
24         for(register int i=1;i<=maxn;++i)
25         {
26             c[i]=d[i]=0;
27         }
28         int kl;
29         for(int i=1;i<=m;++i)
30          for(int j=1;j<=n;++j)
31          {
32              scanf("%d",&a[i][j]);
33              if(a[i][j]==0){kl=i;continue; }
34              d[++cnt]=a[i][j];
35              update(d[cnt],1);
36              num+=cnt-getsum(d[cnt]);
37          }
38         if(n%2==1)
39         {
40             if(num%2==0)printf("YES
");
41             else printf("NO
");
42         }
43         else 
44         {
45             if((m-kl+num)%2==0)printf("YES
");
46             else printf("NO
"); 
47         }
48     }
49 }