Codeforces Round #472 Div. 1

　　A：某个格子被染黑说明该行和该列同时被选中，使用并查集合并，最后看每个集合中是否有未被染黑的格子即可。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 55
char getc(){char c=getchar();while (c!='.'&&c!='#') 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,fa[N<<1],u[N],v[N];
char a[N][N];
void error(){cout<<"No";exit(0);}
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
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+m;i++) fa[i]=i;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
		{
			a[i][j]=getc();
			if (a[i][j]=='#') fa[find(j+n)]=find(i);
		}
	for (int i=1;i<=n+m;i++)
	{
		int p=0,q=0;
		for (int j=1;j<=n;j++)
		if (find(j)==i) u[++p]=j;
		for (int j=1;j<=m;j++)
		if (find(j+n)==i) v[++q]=j;
		for (int x=1;x<=p;x++)
			for (int y=1;y<=q;y++)
			if (a[u[x]][v[y]]=='.') error();
	}
	cout<<"Yes";
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　B：显然E_i和E_j相邻时最优，在固定E_i后E_k越靠后越优，双指针即可。然而我这个弱智硬是写了个分数规划+单调队列。

#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!='.'&&c!='#') 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],q[N];
long double calc(int j,long double k){return a[j]-a[j-1]*k;}
bool check(long double k)
{
	int head=1,tail=0;q[++tail]=2;
	for (int i=3;i<=n;i++)
	{
		while (a[q[head]-1]+m<a[i]&&head<=tail) head++;
		if (head<=tail&&a[i]*(1-k)>=calc(q[head],k)) return 1;
		while (head<=tail&&calc(i,k)<=calc(q[tail],k)) tail--;
		q[++tail]=i;
	}
	return 0;
}
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++) a[i]=read();
	bool flag=0;
	for (int i=1;i<n-1;i++) if (a[i+2]-a[i]<=m) {flag=1;break;}
	if (!flag) {cout<<-1;return 0;}
	long double l=0,r=1,ans;
	for (int i=1;i<=200;i++)
	{
		long double mid=(l+r)/2;
		if (check(mid)) ans=mid,l=mid;
		else r=mid;
	}
	double u=ans;printf("%.9f",u);
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　C：严格小于=all-严格大于-1，所以即要最小化每天的标记数之和。显然其要大于给定的m_i，当然令其为m_i+1即最优。但同时每天最多添加一个标记，所以每个数还得对后一个数-1取max。

#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!='.'&&c!='#') 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],b[N];
ll ans;
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(),ans-=a[i]+1;
	for (int i=1;i<=n;i++) b[i]=max(b[i-1],a[i]+1);
	for (int i=n;i>=1;i--) b[i]=max(b[i],b[i+1]-1);
	for (int i=1;i<=n;i++) ans+=b[i];
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　D：显然到达原点的时间随风速的变化是一个单调函数，且拎出两个函数他们的大小关系也是单调的，于是只需要比较两端点的函数值即可。求出速度两个极限，数二维偏序数量就完了。卡精。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define double long double 
#define N 100010
#define lson tree[k].ch[0]
#define rson tree[k].ch[1]
char getc(){char c=getchar();while (c!='.'&&c!='#') 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,root;
const double eps=1E-11;
ll ans;
struct data
{
	double x,y;
	bool operator <(const data&a) const
	{
		return x<a.x||fabs(x-a.x)<eps&&y>a.y;
	}
}a[N],b[N],c[N]; 
struct data2{double x;int ch[2],p,s;
}tree[N];
void up(int k){tree[k].s=tree[lson].s+tree[rson].s+1;}
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,double x)
{
	if (k==0){k=++cnt;tree[k].x=x;lson=rson=0;tree[k].p=rand();tree[k].s=1;return;}
	tree[k].s++;
	if (tree[k].x>x) {ins(lson,x);if (tree[lson].p>tree[k].p) move(k,0);}
	else {ins(rson,x);if (tree[rson].p>tree[k].p) move(k,1);}
}
int query(int k,double x)
{
	if (k==0) return 0;
	if (tree[k].x+eps<x) return query(rson,x);
	else return query(lson,x)+tree[rson].s+1;
}
void clear(){root=cnt=0;}
void solve(int l,int r)
{
	int t=0;
	for (int i=l;i<=r;i++) t++,c[t].x=-a[i].x/(a[i].y+m),c[t].y=-a[i].x/(a[i].y-m);
	sort(c+1,c+t+1);clear();
	for (int i=1;i<=t;i++)
	{
		ans+=query(root,c[i].y);
		ins(root,c[i].y);
	}
}
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	srand(20020509);
	n=read(),m=read();
	for (int i=1;i<=n;i++) a[i].x=-read(),a[i].y=read();
	for (int i=1;i<=n;i++)
	{
		double x=a[i].x/(a[i].y+m),y=a[i].x/(a[i].y-m);
		a[i]=(data){x,y};
	}
	sort(a+1,a+n+1);
	for (int i=1;i<=n;i++)
	{
		ans+=query(root,a[i].y);
		ins(root,a[i].y);
	}
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　E：确定了哪些重要盒子放在顶部且不计入答案后，剩余的重要盒子看起来应该从大到小放置，但有一些比较麻烦的边界情况比如样例2。不妨将堆出来的塔翻转过来，这样就变成了要求顶部在范围内，问题就变得清新起来。于是先背包出不重要盒子的所有高度，然后将重要盒子从大到小排序，设f[i][j]为前i个重要盒子高度为j时的最优答案，转移显然，初值即仅对存在的不重要盒子集合高度设为0，其他设为-inf。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<bitset>
using namespace std;
#define ll long long
#define N 10010
char getc(){char c=getchar();while (c!='.'&&c!='#') 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,l,r,u,v,a[N],b[N],c[N],f[N],ans;
bitset<N> F;
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	n=read(),l=read(),r=read();
	for (int i=1;i<=n;i++) m+=c[i]=read();
	l=m-l,r=m-r;swap(l,r);
	for (int i=1;i<=n;i++) if (read()) a[++u]=c[i];else b[++v]=c[i];
	F[0]=1;
	for (int i=1;i<=v;i++)
	F|=F<<b[i];
	sort(a+1,a+u+1);reverse(a+1,a+u+1);
	memset(f,200,sizeof(f));for (int i=0;i<=10000;i++) if (F[i]) f[i]=0;
	for (int i=1;i<=u;i++)
		for (int j=10000;j>=a[i];j--)
		f[j]=max(f[j],f[j-a[i]]+(j>=l&&j<=r));
	for (int i=0;i<=10000;i++) ans=max(ans,f[i]);
	cout<<ans;
	return 0;
}