Codeforces Round #510 Div. 2 Virtual Participate记

　　这场打的顺手到不敢相信。如果不是vp的话估计肯定打不到这个成绩。

　　A：最大显然，最小的话每次暴力给最小的+1。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 110
int n,m,a[N],mx,s;
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    n=read(),m=read();
    for (int i=;i<=n;i++) a[i]=read(),s=max(s,a[i]);
    s+=m;
    for (int i=;i<=m;i++)
    {
        int mx=;
        for (int j=;j<=n;j++) if (a[j]<a[mx]) mx=j;
        a[mx]++;
    }
    int mx=;for (int i=;i<=n;i++) mx=max(mx,a[i]);
    cout<<mx<<' '<<s;
    return ;
}

　　B：每种蔬菜集合取最小值，枚举选哪些即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 1010
int n,ans=,f[];
struct data{int x,y;
}a[N];
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    n=read();for (int i=;i<;i++) f[i]=ans;
    for (int i=;i<=n;i++)
    {
        a[i].x=read();
        char c=getchar();
        while (c<'A'||c>'Z') c=getchar();
        while (c>='A'&&c<='Z')
        {
            a[i].y+=<<c-'A';
            c=getchar();
        }
        f[a[i].y]=min(f[a[i].y],a[i].x);
    }
    for (int i=;i<;i++)
        for (int j=;j<;j++)
            for (int k=;k<;k++)
            if ((i|j|k)==) ans=min(ans,f[i]+f[j]+f[k]);
    if (ans==) cout<<-;
    else cout<<ans;
    return ;
}

　　C：显然就是取一些数使乘积最大。讨论一下各种情况。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 200010
int n,a[N];
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    n=read();
    int cnt=;
    for (int i=;i<=n;i++)
    {
        a[i]=read();
        cnt+=a[i]<;
    }
    int last=;
    if (cnt%==)
    {
        int cnt=;
        for (int i=;i<=n;i++)
        if (a[i]==)
        {
            cnt++;
            if (last) printf("%d %d %d\n",,last,i);
            last=i;
        }
        if (last&&cnt<n) printf("%d %d\n",,last);
        last=;
        for (int i=;i<=n;i++)
        if (a[i]!=)
        {
            if (last) printf("%d %d %d\n",,last,i);
            last=i;
        }
    }
    else
    {
        int mx=,cnt=;
        for (int i=;i<=n;i++)
        if (a[i]<&&(!mx||a[i]>a[mx])) mx=i;
        for (int i=;i<=n;i++)
        if (a[i]==) cnt++,printf("%d %d %d\n",,i,mx);
        if (cnt<n-) printf("%d %d\n",,mx);
        last=;
        for (int i=;i<=n;i++)
        if (a[i]!=&&i!=mx)
        {
            if (last) printf("%d %d %d\n",,last,i);
            last=i;
        }
    }
    return ;
}

　　D：前缀和后即每次询问有多少个数与x的和小于y。随便拿个数据结构维护一下，感觉treap比较方便。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 200010
#define ll long long
#define lson tree[k].ch[0]
#define rson tree[k].ch[1]
int n,root,cnt=;
ll m,a[N],ans;
struct data{ll x;int p,ch[],s;
}tree[N];
void up(int k){tree[k].s=tree[lson].s+tree[rson].s+;}
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,ll x)
{
    if (k==){k=++cnt;tree[k].x=x,tree[k].s=,tree[k].p=rand();tree[k].ch[]=tree[k].ch[]=;return;}
    tree[k].s++;
    if (x<tree[k].x) {ins(lson,x);if (tree[lson].p>tree[k].p) move(k,);}
    else {ins(rson,x);if (tree[rson].p>tree[k].p) move(k,);}
}
ll query(int k,ll s)
{
    if (k==) return ;
    if (tree[k].x>s) return query(lson,s);
    else return query(rson,s)+tree[lson].s+;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif
    cin>>n>>m;m--;
    srand();
    for (int i=;i<=n;i++) a[i]=a[i-]+read();
    ins(root,);
    for (int i=;i<=n;i++)
    {
        ans+=i-query(root,a[i]-m-);
        ins(root,a[i]);
    }
    cout<<ans;
    return ;
}

　　E：从小到大倒退期望即可。每次需要知道某格子到其他小于他的格子的距离平方之和，显然横纵坐标可以分别计算，然后展开即可维护。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 1010
#define P 998244353
int n,m,s,sum,Sum[],Sqr[],Cnt[];
int ksm(int a,int k)
{
    int s=;
    for (;k;k>>=,a=1ll*a*a%P) if (k&) s=1ll*s*a%P;
    return s;
}
int trans(int x,int y){return (x-)*m+y;}
struct data{int x,y,v,ans;
}a[N*N];
bool cmp(const data&a,const data&b)
{
    return a.v<b.v;
}
void update(int k,int x,int p)
{
    a[k].ans=(a[k].ans+1ll*x*x%P*Cnt[p]%P+Sqr[p]-2ll*x*Sum[p]%P+P)%P;
}
void ins(int x,int p)
{
    Cnt[p]++;Sum[p]=(Sum[p]+x)%P;Sqr[p]=(Sqr[p]+1ll*x*x%P)%P;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif
    n=read(),m=read();
    for (int i=;i<=n;i++)
        for (int j=;j<=m;j++)
        a[trans(i,j)].x=i,a[trans(i,j)].y=j,a[trans(i,j)].v=read();
    int x=read(),y=read();s=trans(x,y);
    sort(a+,a+n*m+,cmp);
    for (int i=;i<=n*m;i++)
    {
        int t=i;
        while (t<n*m&&a[t+].v==a[i].v) t++;
        for (int j=i;j<=t;j++)
        {
            a[j].ans=sum;
            update(j,a[j].x,),update(j,a[j].y,);
            a[j].ans=1ll*a[j].ans*ksm(i-,P-)%P;
        }
        for (int j=i;j<=t;j++)
        {
            ins(a[j].x,),ins(a[j].y,);
            sum=(sum+a[j].ans)%P;
        }
        i=t;
    }
    for (int i=;i<=n*m;i++)
    if (trans(a[i].x,a[i].y)==s) {cout<<a[i].ans;return ;}
}

　　F：想了半天觉得一点也不可做，于是大力贪心，莫名其妙就过了。设f[i]为i子树内划分集合的最少数量，g[i]为i子树中这些集合的最大深度的最小值。dp时贪心地合并各子树内最大深度最小的那些集合，即把小的合并在一块，大的不管了。并没有想清楚为什么对。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 1000010
int n,m,p[N],degree[N],deep[N],fa[N],f[N],g[N],b[N],root,t=;
struct data{int to,nxt;
}edge[N<<];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
void dp(int k)
{
    bool flag=;
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=fa[k]) flag=,dp(edge[i].to);
    if (flag) f[k]=,g[k]=;
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=fa[k]) f[k]+=f[edge[i].to];
    int cnt=;
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=fa[k]) b[++cnt]=g[edge[i].to]+;
    sort(b+,b+cnt+);
    if (cnt) g[k]=b[];
    for (int i=;i<cnt;i++)
    if (b[i]+b[i+]<=m) f[k]--,g[k]=b[i+];
}
void dfs(int k)
{
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=fa[k])
    {
        fa[edge[i].to]=k;
        deep[edge[i].to]=deep[k]+;
        dfs(edge[i].to);
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif
    n=read(),m=read();
    for (int i=;i<n;i++)
    {
        int x=read(),y=read();
        addedge(x,y),addedge(y,x);
        degree[x]++,degree[y]++;
    }
    for (int i=;i<=n;i++) if (degree[i]>) {root=i;break;}
    dfs(root);
    dp(root);
    cout<<f[root];
    return ;
}

　　result：rank 5