AISing Programming Contest 2019 翻车记

　　A：签到。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
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;
}
int h,w,n;
int main()
{
    n=read(),h=read(),w=read();
    cout<<(n-w+)*(n-h+);
}

　　B：签到*2。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 110
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;
}
int n,a,b,p[N];
int main()
{
    n=read(),a=read(),b=read();
    for (int i=;i<=n;i++) p[i]=read();
    int ans=,x=,y=,z=;
    for (int i=;i<=n;i++)
    {
        if (p[i]<=a) x++;
        if (p[i]>a&&p[i]<=b) y++;
        if (p[i]>b) z++;
    }
    cout<<min(x,min(y,z));
}

　　C：考虑在相邻两异色格间连边，这样同一个连通块的异色点间一定存在合法路径。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 410
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;
}
int n,m,a[N][N],fa[N*N],size[N*N][];
int trans(int x,int y){return (x-)*m+y;}
int wx[]={,,,-},wy[]={,,-,};
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
int main()
{
    n=read(),m=read();
    for (int i=;i<=n;i++)
        for (int j=;j<=m;j++)
        {
            char c=getchar();
            while (c!='.'&&c!='#') c=getchar();
            a[i][j]=c=='.';
        }
    for (int i=;i<=n;i++)
        for (int j=;j<=m;j++)
        {
            fa[trans(i,j)]=trans(i,j);
            size[trans(i,j)][a[i][j]]=;
        }
    for (int i=;i<=n;i++)
        for (int j=;j<=m;j++)
            for (int k=;k<;k++)
            if (i+wx[k]>=&&i+wx[k]<=n&&j+wy[k]>=&&j+wy[k]<=m&&a[i+wx[k]][j+wy[k]]!=a[i][j])
            {
                int p=find(trans(i,j)),q=find(trans(i+wx[k],j+wy[k]));
                if (p!=q)
                {
                    fa[p]=q;
                    size[q][]+=size[p][],size[q][]+=size[p][];
                }
            }
    ll ans=;
    for (int i=;i<=n;i++)
        for (int j=;j<=m;j++)
        if (find(trans(i,j))==trans(i,j)) ans+=1ll*size[trans(i,j)][]*size[trans(i,j)][];
    cout<<ans;
}

　　D：显然分为两个过程，第一个过程中两人取数互不相关，第二个过程中两人从大到小依次取数。二分套二分找到分界点，瞎搞一波前缀和即可。不知道为啥写了一年。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
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;
}
int n,m,a[N];
ll s[N],f[N];
int findmax(int x,int k)
{
    int l=,r=n;
    while (l+<r)
    {
        int mid=l+r>>;
        if (mid+k->n) r=mid;
        else if (x-a[mid]>a[mid+k-]-x) l=mid;
        else r=mid;
    }
    int ans=l;
    for (int i=l+;i<=r;i++)
    if (r+k-<=n&&max(x-a[i],a[i+k-]-x)<max(x-a[ans],a[ans+k-]-x)) ans=i;
    return a[ans+k-];
}//和x最接近的数中取k个后最大的
bool check(int k,int x)
{
    if (k*->n) return ;
    //cout<<k<<' '<<x<<endl;
    //cout<<findmax(x,1)<<' '<<a[n-k+1]<<endl;
    return findmax(x,k-)<a[n-k+];
}
int main()
{
    n=read(),m=read();
    for (int i=;i<=n;i++) a[i]=read();
    sort(a+,a+n+);
    s[]=a[];
    for (int i=;i<=n;i++) s[i]=s[i-]+a[i];
    for (int i=n;i>=;i--) f[i]=f[i+]+a[i];
    for (int i=;i<=m;i++)
    {
        int x=read();
        int l=,r=n,t=;
        while (l<=r)
        {
            int mid=l+r>>;
            if (check(mid,x)) t=mid,l=mid+;
            else r=mid-;
        }
        //cout<<t<<endl;
        ll ans=f[n-t+];int y=n-t*;
        if (y>) ans+=s[y];
        printf("%lld\n",ans);
    }
}

　　E：正常的想法是设f[i][j]为i子树中包含j的连通块权值和为j时最少要割多少边，显然这样不行，于是就反过来，设f[i][j]为i子树中割j条边能使i所在连通块获得的最小权值和（当然割出的除根以外的连通块均需要合法）。同时设g[i]表示i子树中最少割多少条边能使i所在连通块均为正数（前提同上）。转移类似树形背包。莫名其妙写的非常慢感觉非常恶心。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 5010
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;
}
int n,a[N],p[N],g[N],size[N],t;
ll sum[N],f[N][N];
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 dfs(int k,int from)
{
    sum[k]=a[k];
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=from)
    {
        dfs(edge[i].to,k);
        sum[k]+=sum[edge[i].to];
    }
    if (a[k]>)
    {
        g[k]=;
        for (int i=p[k];i;i=edge[i].nxt)
        if (edge[i].to!=from)
        {
            int x=g[edge[i].to];
            for (int j=;j<=n;j++)
            if (f[edge[i].to][j]<) {x=min(x,j+);break;}
            g[k]+=x;
        }
    }
    f[k][]=a[k];
    size[k]=;
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=from)
    {
        for (int j=size[k]+size[edge[i].to];j>=;j--)
        {
            f[k][j]+=sum[edge[i].to];if (j-g[edge[i].to]->=) f[k][j]=min(f[k][j],f[k][j-g[edge[i].to]-]);
            for (int x=max(,j-size[k]);x<=min(j,size[edge[i].to]);x++)
            {
                if (x) f[k][j]=min(f[k][j],f[k][j-x]+f[edge[i].to][x]);
                if (x<j&&f[edge[i].to][x]<) f[k][j]=min(f[k][j],f[k][j-x-]);
            }
        }
        size[k]+=size[edge[i].to];
    }
}
int main()
{
    /*freopen("e.in","r",stdin);
    freopen("e.out","w",stdout);*/
    n=read();
    for (int i=;i<=n;i++) a[i]=read();
    for (int i=;i<n;i++)
    {
        int x=read(),y=read();
        addedge(x,y),addedge(y,x);
    }
    memset(g,,sizeof(g));
    memset(f,,sizeof(f));
    dfs(,);
    /*for (int i=1;i<=n;i++) cout<<a[i]<<' ';cout<<endl;
    for (int i=1;i<=n;i++) cout<<g[i]<<' ';cout<<endl;
    cout<<endl;
    for (int i=1;i<=n;i++)
    {
        for (int j=0;j<n;j++)
        cout<<f[i][j]<<' ';
        cout<<endl;;
    }
    for (int i=1;i<=n;i++) cout<<sum[i]<<' ';cout<<endl;*/
    int ans=g[];
    for (int i=;i<=n;i++)
    if (f[][i]<) {ans=min(ans,i);break;}
    cout<<ans;
    return ;
}

　　为什么算rating的时候不把unrated的去掉啊。result：rank 43 rating +138