标签：efi 判环 n+1 for col stdin ring iostream scan
A：求出该行该列各有多少个比其小的取max，该行该列各有多少个比其大的取max，加起来即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1010
char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,a[N][N],ans[N][N],ans2[N][N],b[N];
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
n=read(),m=read();
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
a[i][j]=read();
for (int i=1;i<=n;i++)
{
for (int j=1;j<=m;j++) b[j]=a[i][j];
sort(b+1,b+m+1);
int t=unique(b+1,b+m+1)-b-1;
for (int j=1;j<=m;j++)
{
int x=lower_bound(b+1,b+t+1,a[i][j])-b;
ans[i][j]=max(ans[i][j],x);
ans2[i][j]=max(ans2[i][j],t-x);
}
}
for (int i=1;i<=m;i++)
{
for (int j=1;j<=n;j++) b[j]=a[j][i];
sort(b+1,b+n+1);
int t=unique(b+1,b+n+1)-b-1;
for (int j=1;j<=n;j++)
{
int x=lower_bound(b+1,b+t+1,a[j][i])-b;
ans[j][i]=max(ans[j][i],x);
ans2[j][i]=max(ans2[j][i],t-x);
}
}
for (int i=1;i<=n;i++)
{
for (int j=1;j<=m;j++)
printf("%d ",ans[i][j]+ans2[i][j]);
printf("\n");
}
return 0;
//NOTICE LONG LONG!!!!!
}
B：kmp求出最长border，贪心的每次利用这个border制造新子串即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1000010
char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,cnt,nxt[N],cnt2;
char a[N],b[N];
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
scanf("%s",a+1);n=strlen(a+1);
scanf("%s",b+1);m=strlen(b+1);
for (int i=1;i<=n;i++) if (a[i]==‘1‘) cnt++;
for (int i=1;i<=m;i++) if (b[i]==‘1‘) cnt2++;
nxt[0]=-1;
for (int i=1;i<=m;i++)
{
int j=nxt[i-1];
while (b[j+1]!=b[i]&&j!=-1) j=nxt[j];
nxt[i]=j+1;
}
if (n<m) {printf("%s",a+1);return 0;}
if (cnt<cnt2||n-cnt<m-cnt2) {printf("%s",a+1);return 0;}
int X=cnt,Y=n-cnt;
printf("%s",b+1);X-=cnt2,Y-=m-cnt2;
cnt=0;for (int i=nxt[m]+1;i<=m;i++) if (b[i]==‘1‘) cnt++;
cnt2=m-nxt[m]-cnt;
while (X>=cnt&&Y>=cnt2)
{
for (int i=nxt[m]+1;i<=m;i++) putchar(b[i]);
X-=cnt,Y-=cnt2;
}
for (int i=1;i<=X;i++) printf("1");
for (int i=1;i<=Y;i++) printf("0");
return 0;
//NOTICE LONG LONG!!!!!
}
C：场上写了1h假题意。终于看懂题之后又被之前假题意的做法带偏了。考虑拆点，建出新图后缩点，跑最长链即为答案。因为由图的特殊性容易发现，若新图中有两个点代表原图中同一个点的不同时刻且其弱联通，其一定处于同一SCC中。所以不会出现多次计算同一个点的贡献的情况。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define N 100010
#define M 54
char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,d,p[N],dfn[N*M],low[N*M],stk[N*M],Set[N*M],P[N*M],nxt[N*M],top,cnt,t,ans,a[N][M];
bool flag[N*M];
struct data{int to,nxt;
}edge[N<<1];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
int id(int n,int m){return (n-1)*d+m;}
void tarjan(int x,int y)
{
int k=id(x,y);
dfn[k]=low[k]=++cnt;
stk[++top]=k;flag[k]=1;
for (int i=p[x];i;i=edge[i].nxt)
if (!dfn[id(edge[i].to,y%d+1)]) tarjan(edge[i].to,y%d+1),low[k]=min(low[k],low[id(edge[i].to,y%d+1)]);
else if (flag[id(edge[i].to,y%d+1)]) low[k]=min(low[k],dfn[id(edge[i].to,y%d+1)]);
if (dfn[k]==low[k])
{
t++;
while (stk[top]!=k)
{
nxt[stk[top]]=P[t];
P[t]=stk[top];
flag[stk[top]]=0;
Set[stk[top]]=t;
top--;
}
flag[k]=0,Set[k]=t,top--;
nxt[k]=P[t];P[t]=k;
}
}
namespace DAG
{
int p[N*M],n,t,f[N*M],value[N*M];
struct data{int to,nxt;}edge[N*M];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
void dp()
{
for (int i=1;i<=n;i++)
{
for (int j=p[i];j;j=edge[j].nxt)
f[i]=max(f[i],f[edge[j].to]);
f[i]+=value[i];
}
}
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
n=read(),m=read(),d=read();
for (int i=1;i<=m;i++)
{
int x=read(),y=read();
addedge(x,y);
}
for (int i=1;i<=n;i++)
for (int j=1;j<=d;j++)
{
char c=getchar();
while (c!=‘0‘&&c!=‘1‘) c=getchar();
a[i][j]=c-‘0‘;
}
t=0;
for (int i=1;i<=n;i++)
for (int j=1;j<=d;j++)
if (!dfn[id(i,j)]) tarjan(i,j);
DAG::n=t;
for (int i=1;i<=n;i++)
for (int x=1;x<=d;x++)
for (int j=p[i];j;j=edge[j].nxt)
if (Set[id(i,x)]!=Set[id(edge[j].to,x%d+1)])
DAG::addedge(Set[id(i,x)],Set[id(edge[j].to,x%d+1)]);
memset(flag,0,sizeof(flag));
for (int i=1;i<=t;i++)
{
for (int j=P[i];j;j=nxt[j])
if (!flag[(j-1)/d+1]&&a[(j-1)/d+1][(j-1)%d+1]) DAG::value[i]++,flag[(j-1)/d+1]=1;
for (int j=P[i];j;j=nxt[j]) flag[(j-1)/d+1]=0;
}
DAG::dp();
cout<<DAG::f[Set[1]];
return 0;
//NOTICE LONG LONG!!!!!
}
D：看到这张图就容易想到floyd判环。于是先拉两个点出来跑一遍，大约能得到一个2(t+d)=t+d+kc。其中k和c都可以确定。这样大概可以二分一下t，但点数说不定不太够。事实上这个时候直接让所有点一起往前跑第一次撞上就是答案了。因为t+d=kc，当前环上点在d的位置，走t步后刚好走到终点。同时由起点出发的点也刚好走了t步到达终点。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
int x=0,f=1;char c=getchar();
while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
char s[12];
int main()
{
#ifndef ONLINE_JUDGE
freopen("b.in","r",stdin);
freopen("b.out","w",stdout);
#endif
int cnt=0;
while (1)
{
cout<<"next 1 2"<<endl;
int x=read();
for (int i=1;i<=x;i++) scanf("%s",s);
cout<<"next 1"<<endl;
x=read();
for (int i=1;i<=x;i++) scanf("%s",s);
if (x==2) break;
}
while (1)
{
cout<<"next 0 1 2 3 4 5 6 7 8 9"<<endl;
int x=read();
for (int i=1;i<=x;i++) scanf("%s",s);
if (x==1) {cout<<"done"<<endl;return 0;}
}
}
剩下的先咕着。
感觉非常可怕的是看完sol不是感叹题好神而是感叹自己傻逼。
靠着偶尔快起来的切傻逼题的手速上了红，事实上还是打铁实力。简直自闭。
小号打的。result：rank 254 rating -19
Codeforces Round #545 Div. 1自闭记
标签：efi 判环 n+1 for col stdin ring iostream scan
原文地址：https://www.cnblogs.com/Gloid/p/10498984.html
