This is probably my most recent one from the mini skirt
B
In fact, some of the entire table Unicom block negate operation can not cross the communication block. Therefore, direct statistics about each block within the digital communication is not an even number not to
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define gt(x,y) ((x)*m+y)
using namespace std;
const int M = 1100000;
int n,m,k,a[M],b[M],d[M],T,r,c,cnt=0,s[M];
char C[M];
bool check(int x,int y)
{
if(x<0 || y<0 || x>=n || y>=m) return 0;
return 1;
}
void dfs(int x,int y)
{
if(!check(x,y)) return ;
if(d[gt(x,y)]) return ;
d[gt(x,y)]=cnt;
dfs(x-r,y); dfs(x+r,y); dfs(x,y-c); dfs(x,y+c);
}
int main()
{
scanf("%d",&T);
for(;T;T--)
{
scanf("%d%d%d%d",&n,&m,&r,&c);
int B=1; cnt=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++) if(!d[gt(i,j)])
{
cnt++; dfs(i,j);
}
for(int i=0;i<n;i++)
{
scanf("\n%s",C);
for(int j=0;j<m;j++) a[gt(i,j)]=C[j]-'0';
}
for(int i=0;i<n;i++)
{
scanf("\n%s",C);
for(int j=0;j<m;j++)
{
k=C[j]-'0';
if(k!=a[gt(i,j)]) s[d[gt(i,j)]]^=1;
}
}
for(int i=1;i<=cnt;i++) if(s[i]) B=0;
if(B) printf("Yes\n");
else printf("No\n");
for(int i=0;i<=n*m;i++) a[i]=b[i]=d[i]=s[i]=0;
}
}C
Built directly map the shortest run
Note NOTE: If you have two buses stop at a station colleagues, among them to even two-way side!
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#define gt(x,y) ((x)*m+y)
using namespace std;
const int M = 2100000;
int t,m,k,b[M],d[M],T,cnt=0,n1,n2,s,t1,t2,head[M],nex[M*3],edge[M*3],ver[M*3],S;
priority_queue<pair<int,int> >q;
struct vv
{
int a,b;
} v[M];
bool cmp(vv a,vv b){return a.a<b.a;}
void add(int x,int y,int z)
{
ver[++cnt]=y, nex[cnt]=head[x], head[x]=cnt; edge[cnt]=z;
}
void dj()
{
for(int i=0;i<=t;i++) d[i]=0x3f3f3f3f, b[i]=0;
while(q.size()) q.pop(); d[S]=0;
q.push(make_pair(0,S));
while(q.size())
{
while(q.size() &&(b[q.top().second])) q.pop();
if(!q.size()) break;
int x=q.top().second; q.pop(); b[x]=1;
for(int i=head[x];i;i=nex[i])
{
int t=ver[i];
if(d[x]+edge[i]<d[t])
{
d[t]=d[x]+edge[i];
q.push(make_pair(-d[t],t));
}
}
}
}
int main()
{
scanf("%d",&T);
for(;T;T--)
{
scanf("%d%d%d%d%d%d",&m,&n1,&n2,&s,&t1,&t2);
t=m*(n1+n2)+2; S=t-1; cnt=0;
for(int i=0;i<m;i++)
{
for(int j=0;j<n1+n2;j++) scanf("%d",&v[j].a), v[j].b=j;
sort(v,v+n1+n2,cmp);
for(int j=0;j<n1+n2-1;j++)
{
add(gt(v[j].b,i),gt(v[j+1].b,i),v[j+1].a-v[j].a);
if(v[j+1].a==v[j].a) add(gt(v[j+1].b,i),gt(v[j].b,i),v[j+1].a-v[j].a);
}
if(i)
{
for(int j=0;j<n1;j++) add(gt(j,i-1),gt(j,i),0);
for(int j=0;j<n2;j++) add(gt(n1+j,i),gt(n1+j,i-1),0);
}
if(i!=s-1) continue;
for(int j=0;j<n1+n2;j++) if(v[j].a>=t1 && v[j].a<=t2) add(S,gt(v[j].b,i),v[j].a-t1);
for(int j=0;j<n1+n2;j++) if(v[j].a>=t1 && v[j].a<=t2) add(gt(v[j].b,i),t,t2-v[j].a);
}
dj();
printf("%d\n",min(d[t],t2-t1));
for(int i=0;i<=cnt;i++) edge[i]=ver[i]=nex[i]=0;
for(int i=0;i<=t;i++) head[i]=0;
}
}D
And n exclusive OR of these two numbers illustrate only the bit value of 1 and n is different
This means that if there is an i n bits, then the two numbers one and only one bit i is 1; otherwise the same as the number two this bit value
Then the difference of two numbers can not exceed m, the maximum can be calculated referred to as a sub-set G n (where Wo is seeking a time-out error to the 15s 
Then you can digit dp it!
\ (f [i] [A ] [B] [G] \) represents the i-th bit in the binary, x is not a limit, y is not b ceiling, xy in which or have not reached the upper limit of g of
Miu Miu transfer method because of what I did not expect, I choose to write the transfer
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long
using namespace std;
const int M = 100001;
int T;
LL n,m,k,f[2][2][2][4],a,b,w,g;
int main()
{
scanf("%d",&T);
for(;T;T--)
{
scanf("%lld%lld%lld%lld",&a,&b,&n,&m); g=0;
LL e=0,S=0;
for(int i=60;i>=0;i--) if((1ll<<i)&n)
{
S+=(1ll<<i);
if(e+(1ll<<i)-(n-S)<=m) g+=(1ll<<i), e+=(1ll<<i);
else e-=(1ll<<i);
}
if(n-g>g || e<0)
{
printf("0\n");
continue;
}
for(int i=60;i>=0;i--) if(g&(1ll<<i))
{
w=(1ll<<i);
break;
}
memset(f,0,sizeof(f));
f[1][1][1][0]=1ll;
for(int i=60;i>=0;i--)
{
int t=i&1;
memset(f[t],0,sizeof(f[t]));
LL r=1ll<<i;
for(int A=0;A<=1;A++)
for(int B=0;B<=1;B++)
for(int k=0;k<=2;k++) if(f[!t][A][B][k])
{
if(n&r)
{
if(w==r)
{
int A1=0,B1=0,k1=0;
if((A && (a&r))||(!A))
{
int B1=0;
if(B && (b&r)==0) B1=1;
f[t][A][B1][1]=f[t][A][B1][1]+f[!t][A][B][k];
}
if((B && (b&r))||(!B))
{
int A1=0;
if(A && (a&r)==0) A1=1;
f[t][A1][B][2]=f[t][A1][B][2]+f[!t][A][B][k];
}
}
else
{
if(((k==1 && (g&r))||k!=1) &&( (A &&(a&r))|| !A))
{
int k1=0;
if(k==2 && ((g&r)==0)) k1=2;
if(k==1 && (g&r)) k1=1;
int B1=0;
if(B && (b&r)==0) B1=1;
f[t][A][B1][k1]=f[t][A][B1][k1]+f[!t][A][B][k];
}
if(((k==2 && (g&r))||k!=2) &&( (B &&(b&r))|| !B))
{
int k1=0;
if(k==1 && ((g&r)==0)) k1=1;
if(k==2 && (g&r)) k1=2;
int A1=0;
if(A && (a&r)==0) A1=1;
f[t][A1][B][k1]=f[t][A1][B][k1]+f[!t][A][B][k];
}
}
}
else
{
int A1=0, B1=0;
if(A && (a&r)==0) A1=1;
if(B && (b&r)==0) B1=1;
f[t][A1][B1][k]=f[t][A1][B1][k]+f[!t][A][B][k];
if((A && (a&r)==0) ||(B && (b&r)==0)) continue;
A1=0, B1=0;
if(A && (a&r)) A1=1;
if(B && (b&r)) B1=1;
f[t][A1][B1][k]=f[t][A1][B1][k]+f[!t][A][B][k];
}
}
}
LL res=0;
for(int i=0;i<=1;i++)
for(int j=0;j<=1;j++)
for(int k=0;k<=2;k++) res=res+f[0][i][j][k];
printf("%lld\n",res);
}
}