Codeforces Global Round 2
A：答案一定包含首位或末位。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 300010
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,a[N];
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
n=read();
for (int i=1;i<=n;i++) a[i]=read();
if (a[1]!=a[n]) cout<<n-1;
else
{
int ans=0;
for (int i=n;i>=1;i--) if (a[i]!=a[1]) ans=max(ans,i-1);
for (int i=1;i<=n;i++) if (a[i]!=a[n]) ans=max(ans,n-i);
cout<<ans;
}
return 0;
//NOTICE LONG LONG!!!!!
}
B：二分答案，大的匹配大的即可。
#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,h,a[N],b[N];
bool check(int k)
{
for (int i=1;i<=k;i++) b[i]=a[i];
sort(b+1,b+k+1);
ll s=0;
for (int i=k;i>0;i-=2) s+=b[i];
return s<=h;
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
n=read(),h=read();
for (int i=1;i<=n;i++) a[i]=read();
int l=1,r=n,ans=1;
while (l<=r)
{
int mid=l+r>>1;
if (check(mid)) ans=mid,l=mid+1;
else r=mid-1;
}
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
C：显然将两矩阵xor一下后相当于将矩阵变为全0矩阵，大胆猜想要求每行每列都有偶数个1即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 510
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],b[N][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++)
a[i][j]^=read();
for (int i=1;i<=n;i++)
{
int s=0;
for (int j=1;j<=m;j++)
s^=a[i][j];
if (s&1) {cout<<"NO";return 0;}
}
for (int i=1;i<=m;i++)
{
int s=0;
for (int j=1;j<=n;j++)
s^=a[j][i];
if (s&1) {cout<<"NO";return 0;}
}
cout<<"YES";
return 0;
//NOTICE LONG LONG!!!!!
}
D：显然询问的答案只与长度有关，离线按长度从小到大排序，相当于一个不断合并间隙的过程，将间隙长度排序记录已经有多少个被填满即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
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,q;
ll a[N],ans[N],d[N];
struct data
{
ll k,i;
bool operator <(const data&a) const
{
return k<a.k;
}
}b[N];
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
n=read();
for (int i=1;i<=n;i++) scanf("%I64d",&a[i]);
sort(a+1,a+n+1);
for (int i=2;i<=n;i++) d[i]=a[i]-a[i-1];
sort(d+2,d+n+1);
q=read();
for (int i=1;i<=q;i++)
{
ll l,r;scanf("%I64d%I64d",&l,&r);
b[i].k=r-l+1;b[i].i=i;
}
sort(b+1,b+q+1);
int cnt=n,x=1;
for (int i=1;i<=q;i++)
{
ans[b[i].i]=ans[b[i-1].i];
while (x<n&&d[x+1]<=b[i].k) ans[b[i].i]-=b[i-1].k,x++,cnt--,ans[b[i].i]+=d[x];
ans[b[i].i]+=1ll*cnt*(b[i].k-b[i-1].k);
}
for (int i=1;i<=q;i++) printf("%I64d ",ans[i]);
return 0;
//NOTICE LONG LONG!!!!!
}
E：显然三角形一定等腰且腰为最长边，瞎贪心，从小到大考虑边长，让其先尽量与之前的短边组成三角形，若有剩余则内部消化，最后留下的边留给更长的边匹配。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 300010
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,a[N];
ll ans,x=0;
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
n=read();
for (int i=1;i<=n;i++) a[i]=read();
for (int i=1;i<=n;i++)
{
if (a[i]>=x*2)
{
ans+=x,a[i]-=x*2,x=0;
ans+=a[i]/3,x=a[i]%3;
}
else
{
ans+=a[i]/2,x-=a[i]/2,x+=a[i]%2;
}
}
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
F：考虑对于单次询问怎么做，显然有dpf[i][0/1]表示在i连向父亲的这条边是否切的情况下，i子树内所有点都满足条件的最小代价，转移将儿子按f[][0]-f[][1]排下序即可。注意到度数>=x的点的个数是O(n/x)的，所以对于所有询问，需要考虑的点的总数只有O(nlogn)个。但同时这些点还会与度数<x的点相连，这些边也可以提供贡献，这一部分可以用平衡树维护，dp时在平衡树上二分决定删除哪些边即可。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 250010
#define inf 10000000000000000ll
#define lson tree[k].ch[0]
#define rson tree[k].ch[1]
mt19937 rnd(20020509);
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,p[N],degree[N],root[N],id[N],t,cnt;
bool vis[N];
ll f[N][2],S[N],a[N];
struct data{int to,nxt,len;
}edge[N<<1];
struct data2{int ch[2],p,size,x;ll sum;
}tree[N<<1];
void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;}
bool cmp(const int&a,const int&b)
{
return degree[a]<degree[b];
}
ll query(int k,int x,int s)
{
if (x==0) return 0;
if (k==0) return x>s?inf:S[x];
if (tree[lson].size>=x) return query(lson,x,s);
if (x-tree[lson].size>s) return query(rson,x-tree[lson].size-1,s)+tree[lson].sum+tree[k].x;
ll t=tree[lson].sum+tree[k].x+S[x-tree[lson].size-1];
if (tree[k].x<=a[x-tree[lson].size]) return t=min(t,tree[lson].sum+tree[k].x+query(rson,x-tree[lson].size-1,s));
else return t=min(t,query(lson,x,s));
return t;
}
void up(int k)
{
tree[k].size=tree[lson].size+tree[rson].size+1;
tree[k].sum=tree[lson].sum+tree[rson].sum+tree[k].x;
}
void move(int &k,int p)
{
int t=tree[k].ch[p];
tree[k].ch[p]=tree[t].ch[!p],tree[t].ch[!p]=k,up(k),up(t),k=t;
}
void ins(int &k,int x)
{
if (k==0) {k=++cnt;tree[k].p=rnd(),tree[k].size=1,tree[k].sum=tree[k].x=x;return;}
if (tree[k].x<x) {ins(rson,x);if (tree[rson].p>tree[k].p) move(k,1);}
else {ins(lson,x);if (tree[lson].p>tree[k].p) move(k,0);}
up(k);
}
void dfs(int k,int from)
{
vis[k]=1;int x=0;bool isroot=1;f[k][0]=f[k][1]=0;
int last=0;
for (int i=p[k];i;i=edge[i].nxt)
if (degree[edge[i].to]>_)
{
if (!vis[edge[i].to]) dfs(edge[i].to,k),f[k][0]+=f[edge[i].to][1],f[k][1]+=f[edge[i].to][1];
else f[k][0]+=edge[i].len,isroot=0;
last=i;
}
else if (last) edge[last].nxt=edge[i].nxt;else p[k]=edge[i].nxt;
for (int i=p[k];i;i=edge[i].nxt)
if (degree[edge[i].to]>_&&edge[i].to!=from) a[++x]=f[edge[i].to][0]-f[edge[i].to][1];
sort(a+1,a+x+1);
for (int i=1;i<=x;i++) S[i]=S[i-1]+a[i];
for (int i=x;i>=0;i--)
if (a[i]<=0)
{
if (degree[k]-i<=_) f[k][1]+=S[i];
else f[k][1]+=query(root[k],degree[k]-_,x);
break;
}
if (!isroot)
{
for (int i=x;i>=0;i--)
if (a[i]<=0)
{
if (degree[k]-1-i<=_) f[k][0]+=S[i];
else f[k][0]+=query(root[k],degree[k]-1-_,x);
break;
}
}
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
n=read();
for (int i=1;i<n;i++)
{
int x=read(),y=read(),z=read();
addedge(x,y,z),addedge(y,x,z);
degree[x]++,degree[y]++;
}
for (int i=1;i<=n;i++) id[i]=i;
sort(id+1,id+n+1,cmp);
int x=1;
for (_=0;_<n;_++)
{
while (x<=n&°ree[id[x]]<=_)
{
for (int j=p[id[x]];j;j=edge[j].nxt)
ins(root[edge[j].to],edge[j].len);
x++;
}
for (int j=x;j<=n;j++) vis[id[j]]=0;
ll ans=0;
for (int j=x;j<=n;j++)
if (!vis[id[j]])
{
dfs(id[j],id[j]);
ans+=f[id[j]][1];
}
//for (int i=1;i<=n;i++) cout<<f[i][0]<<' '<<f[i][1]<<endl;
printf("%I64d ",ans);
}
return 0;
//NOTICE LONG LONG!!!!!
}
G、H：咕咕咕
小号打的。这种场真是不管打成什么样都能涨分。result：rank 310 rating +7
