【POJ2893&HDOJ6620】M × N Puzzle（n*m数码判定）

题意：给定一个n*m的矩阵，其中不重复地填【0，n*m-1】，问是否能通过有限步数将0移到右下角

n,m<=1e3

思路：结论题 当板子了

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 //typedef pair<ll,ll>P;
 #define N  1000010
 #define M  200010
 #define fi first
 #define se second
 #define MP make_pair
 #define pb push_back
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const int MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       int INF=1e9;
       int inf=0x7fffffff;
       int dx[]={-,,,};
       int dy[]={,,-,};

       int t[N],mx;

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 int query(int x)
 {
     int res=;
     while(x<=mx)
     {
         res^=t[x];
         x+=lowbit(x);
     }
     return res;
 }

 void add(int x)
 {
     while(x)
     {
         t[x]^=;
         x-=lowbit(x);
     }
 }

 int main()
 {
     int n,m;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         if(!n) break;
         mx=n*m;
         rep(i,,mx) t[i]=;
         int s=,ans=,k=;
         rep(i,,n)
          rep(j,,m)
          {
              s++;
              int x=read();
              if(!x) k=n-i;
               else
               {
                   ans^=query(x);
                   add(x);
               }
          }
         if(m&) k=;
         if((ans+k)%==) printf("Yes\n");
          else printf("No\n");
     }
     return ;
 }