2020.11.19 刷题记录

1 tran

题意：给出一个图，求从任一位置射入光束后光束存在的最长时间与射入方向。若存在多种时间相同的，输出字典序最小的方案。
题解：比较water的大模拟，题意看对很快能想出来

CODE

 
#include
using namespace std;
#define re register
#define LL long long
#define DB double
#define il inline
#define For(x,a,b) for(re int x=a;x<=b;x++)
#define For2(x,a,b) for(re int x=a;x>=b;x--)
#define LFor(x,a,b) for(re LL x=a;x<=b;x++)
#define LFor2(x,a,b) for(re LL x=a;x>=b;x--)
#define Abs(x) ((x>0)? x:-x)
#define INF 100000000
#define pii pair
#define fi first
#define se second
#define mabs(x) ((abs(x))==(0)? (INF):(abs(x)))
int gi()
{
    int res=0,fh=1;char ch=getchar();
    while((ch>'9'||ch<'0')&&ch!='-') ch=getchar();
    if(ch=='-') fh=-1,ch=getchar();
    while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();
    return fh*res;
}
LL gl()
{
    LL res=0,fh=1;char ch=getchar();
    while((ch>'9'||ch<'0')&&ch!='-') ch=getchar();
    if(ch=='-') fh=-1,ch=getchar();
    while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();
    return fh*res;
}
char mp[4]={'U','D','L','R'};
int n,m,a[505][505];
bool mrk[505][505][4][4];
int Time,ans[10];
int solve(pii POS,pii D)
{
	int x=POS.fi,y=POS.se,dx=D.fi,dy=D.se;
	if(x<1 || x>n || y<1 || y>m || a[x][y]==2)
		return Time;
	if(mrk[x][y][dx+1][dy+1])
			return INF;
	mrk[x][y][dx+1][dy+1]=true;
	Time++;
	if(a[x][y]==0) return solve(pii(x+dx,y+dy),D);
	int ndx=dy*a[x][y],ndy=dx*a[x][y];
	return solve(pii(x+ndx,y+ndy),pii(ndx,ndy));
}
void init()
{
	Time=0;
	memset(mrk,0,sizeof(mrk));
}
int main()

{

//	freopen("tran.in","r",stdin);

//	freopen("tran.out","w",stdout);

ios::sync_with_stdio(false);

cin>>n>>m;

For(i,1,n)

For(j,1,m)

{

char t;

cin>>t;

if(t'/')a[i][j]=-1;

else if(t'\') a[i][j]=1;

else if(t'.') a[i][j]=0;

else a[i][j]=2;

}

pii ray;

cin>>ray.fi>>ray.se;

init();

ans[0]=solve(ray,pii(-1,0));

init();

ans[1]=solve(ray,pii(1,0));

init();

ans[2]=solve(ray,pii(0,-1));

init();

ans[3]=solve(ray,pii(0,1));

int ind=-1;

For(i,0,3) if(ind-1 || ans[ind]<ans[i])ind=i;

cout<<mp[ind]<<endl;

if(ans[ind]>=INF) cout<<"Voyager"<<endl;

else cout<<ans[ind]<<endl;

return 0;

}

2 week

题意：有两个数x,y,进行n次操作，对于第i次,有如下两种操作类型可任选一：

1. x+=a[i],y-=b[i]
2. x-=c[i],y+=d[i]

求最终可以得到的\(x*y\)的最大值
题解：送分题。。。直接枚举\(2^n\)种方案，记录个最大值就行。然后记得开long long。

CODE

 
#include
using namespace std;
#define re register
#define LL long long
#define DB double
#define il inline
#define For(x,a,b) for(re int x=a;x<=b;x++)
#define For2(x,a,b) for(re int x=a;x>=b;x--)
#define LFor(x,a,b) for(re LL x=a;x<=b;x++)
#define LFor2(x,a,b) for(re LL x=a;x>=b;x--)
#define Abs(x) ((x>0)? x:-x)
#define INF 1000000000009
#define mabs(x) ((abs(x))==(0)? (INF):(abs(x)))
int gi()
{
    int res=0,fh=1;char ch=getchar();
    while((ch>'9'||ch<'0')&&ch!='-') ch=getchar();
    if(ch=='-') fh=-1,ch=getchar();
    while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();
    return fh*res;
}
LL gl()
{
    LL res=0,fh=1;char ch=getchar();
    while((ch>'9'||ch<'0')&&ch!='-') ch=getchar();
    if(ch=='-') fh=-1,ch=getchar();
    while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();
    return fh*res;
}
int n,a[20],b[20],c[20],d[20];
LL ans,X,Y;
void dfs(int pos) 
{
    if (pos>n) 
	{
        ans=max(ans,X*Y);
        return ;
    }
    LL t1=X,t2=Y;
    X+=a[pos],Y=max((LL)0,Y-b[pos]);
    dfs(pos+1);
    X=t1,Y=t2;
    Y+=c[pos],X=max((LL)0,X-d[pos]);
    dfs(pos+1);
    X=t1,Y=t2;
}
int main()
{
    freopen("week.in","r",stdin);
    freopen("week.out","w",stdout);
    ans=X=Y=0;
    n=gi();
    For(i,1,n) a[i]=gi(),b[i]=gi(),c[i]=gi(),d[i]=gi();
    dfs(1);
    printf("%lld\n",ans);
    return 0;
}
  

3 duty

题意：有一个\(n*m\)的黑白方格图，每次询问其中子矩形内部的黑色连通块个数，保证任意两个联通的黑格间仅有一条路径。
题解：由上面加粗的部分，我们可以知道此图能构造出一张DAG，那么联通块=点数-边数。
二位前缀和维护一下点数和边数即可。注意每次询问考虑的是矩形内部的连通块。

CODE

#include
using namespace std;
#define re register
#define LL long long
#define DB double
#define il inline
#define For(x,a,b) for(re int x=a;x<=b;x++)
#define For2(x,a,b) for(re int x=a;x>=b;x--)
#define LFor(x,a,b) for(re LL x=a;x<=b;x++)
#define LFor2(x,a,b) for(re LL x=a;x>=b;x--)
#define Abs(x) ((x>0)? x:-x)
#define INF 1000000000009
#define mabs(x) ((abs(x))==(0)? (INF):(abs(x)))
int gi()
{
    int res=0,fh=1;char ch=getchar();
    while((ch>'9'||ch<'0')&&ch!='-') ch=getchar();
    if(ch=='-') fh=-1,ch=getchar();
    while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();
    return fh*res;
}
LL gl()
{
    LL res=0,fh=1;char ch=getchar();
    while((ch>'9'||ch<'0')&&ch!='-') ch=getchar();
    if(ch=='-') fh=-1,ch=getchar();
    while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();
    return fh*res;
}
int n,m,q;
int pre_P[2005][2005],pre_E[2005][2005];
int pa[2005][2005],pb[2005][2005];
char s[2005][2005];
int main()

{

n=gi(),m=gi(),q=gi();

For(i,1,n)

scanf("%s",s[i]+1);

For(i,1,n)

For(j,1,m)

{

pre_P[i][j]=pre_P[i-1][j]+pre_P[i][j-1]-pre_P[i-1][j-1]+(s[i][j]'1'? 1:0);

pre_E[i][j]=pre_E[i-1][j]+pre_E[i][j-1]-pre_E[i-1][j-1]+(s[i][j]'1'? (s[i][j]s[i][j-1]? 1:0)+(s[i][j]s[i-1][j]? 1:0):0);

pa[i][j]=pa[i-1][j]+(s[i][j]'1'&&s[i][j]s[i-1][j]? 1:0);

pb[i][j]=pb[i][j-1]+(s[i][j]'1'&&s[i][j]s[i][j-1]? 1:0);

}

//	printf("%d\n",pre_E[3][3]);

re int np,ne,x,y,xx,yy;

For(i,1,q)

{

x=gi(),y=gi(),xx=gi(),yy=gi();

np=pre_P[xx][yy]-pre_P[xx][y-1]-pre_P[x-1][yy]+pre_P[x-1][y-1];

ne=pre_E[xx][yy]-pre_E[xx][y]-pre_E[x][yy]+pre_E[x][y]+pa[xx][y]-pa[x][y]+pb[x][yy]-pb[x][y];

printf("%d\n",np-ne);

}

return 0;

}

/*

5 5 6

11010

01110

10101

11101

01010

1 1 5 5

1 2 4 5

2 3 3 4

3 3 3 3

3 1 3 5

1 1 3 4
3 4 4

1101

0110

1101

1 1 3 4

1 1 3 1

2 2 3 4

1 2 2 4

*/

4 fly

题意：(偷懒)

题解：进行仔细的长时间的分析研究(雾)，发现我们要求的是逆序对数，但瞄一眼空间限制，32MB，在MLE的面前我选择妥协。
继续思考，发现a很小，于是从a入手，发现\(x_i\)构成一个等差数列，于是只要算小于a的逆序对数，剩余的可直接递推计算出来。

CODE

#include
using namespace std;
#define re register
#define LL long long
#define DB double
#define il inline
#define For(x,a,b) for(re int x=a;x<=b;x++)
#define For2(x,a,b) for(re int x=a;x>=b;x--)
#define LFor(x,a,b) for(re LL x=a;x<=b;x++)
#define LFor2(x,a,b) for(re LL x=a;x>=b;x--)
#define Abs(x) ((x>0)? x:-x)
#define INF 1000000000009
#define mabs(x) ((abs(x))==(0)? (INF):(abs(x)))
int gi()
{
	int res=0,fh=1;char ch=getchar();
	while((ch>'9'||ch<'0')&&ch!='-') ch=getchar();
	if(ch=='-') fh=-1,ch=getchar();
	while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();
	return fh*res;
}
LL gl()
{
	LL res=0,fh=1;char ch=getchar();
	while((ch>'9'||ch<'0')&&ch!='-') ch=getchar();
	if(ch=='-') fh=-1,ch=getchar();
	while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();
	return fh*res;
}
int n,x,a,mod;
int tr[500005];
int flag,pls,cut;
LL ans;
int lowbit(int x){return x&-x;}
void change(int x)
{
	for(int i=x;i<=a;i+=lowbit(i))
		tr[i]++;
}
int ask(int x)
{
	int res=0;
	for(int i=x;i;i-=lowbit(i))
		res+=tr[i];
	return res+1;
}
int main()
{
	freopen("fly.in","r",stdin);
	freopen("fly.out","w",stdout);
	n=gi(),x=gi(),a=gi(),mod=gi();
	if(x