由於第二場太水，故不放上來
題目：見2013-5 day1
A
顯然是網絡流，如何建圖是關鍵
我的想法是S->1->2->T，對所有的1拆點
這樣是錯的，我當場wa掉
錯誤的原因我還沒見過——部分流量對整體流量的影響問題，也就是說部分流量是沒用的，我們只關心整體的時候maxflow可能是多個部分的
感覺這個錯誤很難發現，只有調試大數據的時候纔會發現
我說不清楚，看代碼就知道，尤其是我最後判斷flow==2的時候很明顯
代碼（WA）
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
#define rep(i,l,r) for(int i=(l),_=(r);i<=_;i++)
#define per(i,r,l) for(int i=(r),_=(l);i>=_;i--)
#define MS(arr,x) memset(arr,x,sizeof(arr))
#define INE(i,u) for(int i=head[u];~i;i=e[i].next)
#define LL long long
inline const int read()
{int r=0,k=1;char c=getchar();for(;c<'0'||c>'9';c=getchar())if(c=='-')k=-1;
for(;c>='0'&&c<='9';c=getchar())r=r*10+c-'0';return k*r;}
////////////////////////////////////////////////
const int inf=0x3f3f3f3f;
const int N=250000;
const int M=250000;
int n,m;
int A[505][505],B[505][505];
struct edge{int v,flow,cap,next;}e[M];
int head[N],k;
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
int S,T,cnt;
int d[N],q[N],cur[N];
////////////////////////////////////////////////
void adde(int u,int v,int f){e[k]=(edge){v,0,f,head[u]};head[u]=k++;}
void ins(int u,int v,int f1,int f2){adde(u,v,f1);adde(v,u,f2);}
int F(int i){return e[i].cap-e[i].flow;}
bool bfs()
{
memcpy(cur,head,sizeof(cur)); MS(d,-1); d[q[0]=S]=0;
for(int u,v,l=0,r=1;l<r;)
INE(i,u=q[l++]) if(F(i) && d[v=e[i].v]<0)
{
q[r++]=v; d[v]=d[u]+1;
if(v == T) return 1;
}
return 0;
}
int dfs(int u,int a)
{
if(u == T) return a;
for(int &i=cur[u];~i;i=e[i].next) if(F(i) && d[e[i].v] == d[u]+1)
if(int t=dfs(e[i].v,min(a,F(i))))
return e[i].flow+=t,e[i^1].flow-=t,t;
return 0;
}
int dinic()
{
int f=0;
while(bfs()) while(int t=dfs(S,inf)) f+=t;
return f;
}
////////////////////////////////////////////////
void input()
{
MS(head,-1);
n=read(); m=read();
rep(i,1,n) rep(j,1,m) A[i][j]=read();
}
void solve()
{
S=0;
rep(i,1,n) rep(j,1,m) if(A[i][j] == 2)
{
B[i][j] = ++cnt;
}
rep(i,1,n) rep(j,1,m) if(A[i][j] == 1)
{
B[i][j] = ++cnt; cnt+=2;
ins(S,B[i][j],2,0);
ins(B[i][j],B[i][j]+1,1,0);
ins(B[i][j],B[i][j]+2,1,0);
rep(k,0,1)
{
int nx=i+dx[k],ny=j+dy[k];
if(nx < 1 || nx > n || ny < 1 || ny > m) continue;
if(A[nx][ny] != 2) continue;
ins(B[i][j]+1,B[nx][ny],1,0);
}
rep(k,2,3)
{
int nx=i+dx[k],ny=j+dy[k];
if(nx < 1 || nx > n || ny < 1 || ny > m) continue;
if(A[nx][ny] != 2) continue;
ins(B[i][j]+2,B[nx][ny],1,0);
}
}
//rep(i,1,n) {rep(j,1,m) printf("%d ",B[i][j]); puts("");}
T=++cnt;
rep(i,1,n) rep(j,1,m) if(A[i][j] == 2)
{
ins(B[i][j],T,1,0);
}
dinic();
int ans=0;
INE(i,S) if(e[i].flow == 2) ans++;
printf("%d\n",ans);
}
////////////////////////////////////////////////
int main()
{
freopen("A.in","r",stdin); freopen("A.out","w",stdout);
input(),solve();
return 0;
}正解是把所有的1看成橋，212看作一條鏈
有這麼個性質
一條鏈上的兩個2所在的行的奇偶性不同
所以考慮對行的奇偶分類討論
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
#define rep(i,l,r) for(int i=(l),_=(r);i<=_;i++)
#define per(i,r,l) for(int i=(r),_=(l);i>=_;i--)
#define MS(arr,x) memset(arr,x,sizeof(arr))
#define INE(i,u) for(int i=head[u];~i;i=e[i].next)
#define LL long long
inline const int read()
{int r=0,k=1;char c=getchar();for(;c<'0'||c>'9';c=getchar())if(c=='-')k=-1;
for(;c>='0'&&c<='9';c=getchar())r=r*10+c-'0';return k*r;}
////////////////////////////////////////////////
const int inf=0x3f3f3f3f;
const int N=250000;
const int M=250000;
int n,m;
int A[505][505],B[505][505];
struct edge{int v,flow,cap,next;}e[M];
int head[N],k;
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
int S,T,cnt;
int d[N],q[N],cur[N];
////////////////////////////////////////////////
void adde(int u,int v,int f){e[k]=(edge){v,0,f,head[u]};head[u]=k++;}
void ins(int u,int v,int f1,int f2){adde(u,v,f1);adde(v,u,f2);}
int F(int i){return e[i].cap-e[i].flow;}
bool bfs()
{
memcpy(cur,head,sizeof(cur)); MS(d,-1); d[q[0]=S]=0;
for(int u,v,l=0,r=1;l<r;)
INE(i,u=q[l++]) if(F(i) && d[v=e[i].v]<0)
{
q[r++]=v; d[v]=d[u]+1;
if(v == T) return 1;
}
return 0;
}
int dfs(int u,int a)
{
if(u == T) return a;
for(int &i=cur[u];~i;i=e[i].next) if(F(i) && d[e[i].v] == d[u]+1)
if(int t=dfs(e[i].v,min(a,F(i))))
return e[i].flow+=t,e[i^1].flow-=t,t;
return 0;
}
int dinic()
{
int f=0;
while(bfs()) while(int t=dfs(S,inf)) f+=t;
return f;
}
////////////////////////////////////////////////
void input()
{
MS(head,-1);
n=read(); m=read();
rep(i,1,n) rep(j,1,m) A[i][j]=read();
}
void solve()
{
S=0;
rep(i,1,n) rep(j,1,m) if(A[i][j] == 2)
{
B[i][j] = ++cnt;
}
rep(i,1,n) rep(j,1,m) if(A[i][j] == 1)
{
B[i][j] = ++cnt; ++cnt;
ins(B[i][j],B[i][j]+1,1,0);
if(i&1)
{
rep(k,0,1)
{
int nx=i+dx[k],ny=j+dy[k];
if(nx < 1 || nx > n || ny < 1 || ny > m) continue;
if(A[nx][ny] != 2) continue;
ins(B[nx][ny],B[i][j],1,0);
}
rep(k,2,3)
{
int nx=i+dx[k],ny=j+dy[k];
if(nx < 1 || nx > n || ny < 1 || ny > m) continue;
if(A[nx][ny] != 2) continue;
ins(B[i][j]+1,B[nx][ny],1,0);
}
}
else
{
rep(k,2,3)
{
int nx=i+dx[k],ny=j+dy[k];
if(nx < 1 || nx > n || ny < 1 || ny > m) continue;
if(A[nx][ny] != 2) continue;
ins(B[nx][ny],B[i][j],1,0);
}
rep(k,0,1)
{
int nx=i+dx[k],ny=j+dy[k];
if(nx < 1 || nx > n || ny < 1 || ny > m) continue;
if(A[nx][ny] != 2) continue;
ins(B[i][j]+1,B[nx][ny],1,0);
}
}
}
//rep(i,1,n) {rep(j,1,m) printf("%d ",B[i][j]); puts("");}
T=++cnt;
rep(i,1,n) rep(j,1,m) if(A[i][j] == 2)
{
if(i&1)
{
ins(S,B[i][j],1,0);
}
else
{
ins(B[i][j],T,1,0);
}
}
printf("%d\n",dinic());
}
////////////////////////////////////////////////
int main()
{
freopen("A.in","r",stdin); freopen("A.out","w",stdout);
input(),solve();
return 0;
}B
紅果果的枚舉gcd
a b d^2 = y1 + y2 d + y3 a b d
1）y1 > 0
d | y1
枚舉 d
則a b = (y1 + y2 d) / (d^2 - y3 d)
2）y1 = 0 && y2 > 0
a b d = y2 + y3 a b
a b | y2
3）y1 = y2 = 0 && y3 > 0
a b d^2 = y3 a b d
d = y3
ans = -1
4）y1 = y2 = y3 = 0
a b d^2 = 0
ans = 0
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define rep(i,l,r) for(int i=(l),_=(r);i<=_;i++)
#define per(i,r,l) for(int i=(r),_=(l);i>=_;i--)
#define MS(arr,x) memset(arr,x,sizeof(arr))
#define INE(i,u) for(int i=head[u];~i;i=e[i].next)
#define LL long long
inline const int read()
{int r=0,k=1;char c=getchar();for(;c<'0'||c>'9';c=getchar())if(c=='-')k=-1;
for(;c>='0'&&c<='9';c=getchar())r=r*10+c-'0';return k*r;}
////////////////////////////////////////////////
const LL mod=1000000007;
LL y1,y2,y3;
LL ans=0;
////////////////////////////////////////////////
void cal1(LL ab)
{
int cnt=0;
for(int a=2;a*a<=ab;a++) if(ab%a==0)
{
cnt++;
while(ab%a==0) ab/=a;
if(ab==1) break;
}
if(ab>1) cnt++;
ans=ans+(1LL<<cnt)%mod;
if(ans>=mod) ans-=mod;
}
void cal2(LL d)
{
if(d - y3 <= 0) return;
if((y1 + y2 * d) % (d * d - y3 * d) != 0) return;
cal1((y1 + y2 * d) / (d * d - y3 * d));
}
////////////////////////////////////////////////
void input()
{
cin>>y1>>y2>>y3;
}
void solve()
{
ans=0;
if(y1 == 0)
{
if(y2 == 0)
{
if(y3 == 0) ans=0;
else ans=-1;
}
else
{
for(int i=1;i*i<=y2;i++) if(y2%i == 0)
{
cal1(i);
if(i*i == y2) break;
cal1(y2/i);
}
}
}
else
{
for(int i=1;i*i<=y1;i++) if(y1%i == 0)
{
cal2(i);
if(i*i == y1) break;
cal2(y1/i);
}
}
cout<<ans<<endl;
}
////////////////////////////////////////////////
int main()
{
freopen("B.in","r",stdin); freopen("B.out","w",stdout);
rep(i,1,read())
input(),solve();
return 0;
}C