Dhaka2011

Dhaka2011

A - Binary Matrix

题目描述:有一个\(n \times m\)\(01\)矩阵,这一矩阵第一行和最后一行是相邻的,第一列和最后一列是相邻的,现在每次可以交换相邻的两个位置的数(四相邻),问最少多少次操作使得每一行的\(1\)的个数相同,每一列的\(1\)的个数相同,如果不行,则最少多少次操作使得每一行的\(1\)的个数相同,如果不行,则最少多少次操作使得每一列的\(1\)的个数相同,如果不行,则输出无解。

solution
行和列是独立的,因此可以分开做,由于矩阵第一行和最后一行是相邻的,因此要枚举第一行,然后按顺序填补,多则出,少则进,然后取最小值即可。

时间复杂度:\(O(n^2+m^2)\)

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const LL inf=1LL<<60;
const int maxn=1010;

int n, m, total;
int row[maxn], col[maxn];

void read()
{
	scanf("%d%d", &n, &m);
	for (int i=1; i<=n; ++i) row[i]=0;
	for (int i=1; i<=m; ++i) col[i]=0;
	for (int i=1; i<=n; ++i)
		for (int j=1; j<=m; ++j)
		{
			int x;
			scanf("%1d", &x);
			row[i]+=x; col[j]+=x;
		}
}
LL work(int n, int *a)
{
	LL ans=inf;
	int each=total/n;
	for (int i=1; i<=n; ++i)
	{
		int rest=0;
		LL s=0;
		int cur=i;
		for (int j=1; j<=n; ++j, cur=(cur==n? 1:cur+1))
		{
			if (a[cur]>each)
			{
				if (rest<0)
				{
					int used=min(-rest, a[cur]-each);
					s+=used*j; rest+=a[cur]-each;
					s-=(a[cur]-each-used)*j;
				}
				else { s-=(a[cur]-each)*j; rest+=a[cur]-each; }
			}
			else
			{
				if (rest>0)
				{
					int used=min(rest, each-a[cur]);
					s+=used*j; rest-=each-a[cur];
					s-=(each-a[cur]-used)*j;
				}
				else { s-=(each-a[cur])*j; rest-=each-a[cur]; }
			}
		}
		ans=min(ans, s);
	}
	return ans;
}
void solve()
{
	total=0;
	for (int i=1; i<=n; ++i) total+=row[i];
	if (total%n==0 && total%m==0) printf("both ");
	else if (total%n==0) printf("row ");
	else if (total%m==0) printf("column ");
	else { puts("impossible"); return; }

	LL ans=0;
	if (total%n==0) ans+=work(n, row);
	if (total%m==0) ans+=work(m, col);
	printf("%lld\n", ans);
}
int main()
{
	int casesum;
	scanf("%d", &casesum);
	for (int i=1; i<=casesum; ++i)
	{
		printf("Case %d: ", i);
		read();
		solve();
	}
	return 0;
}

B - Candles

solution

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

int n;
int a[20], idx[1<<10];
bool flag[1<<10][210];
LL num[1<<10];

bool cmp(int b, int c)
{
	return num[b]<num[c];
}
void init()
{
	for (int sett=0; sett<1<<10; ++sett)
	{
		int cnt=0;
		for (int i=0; i<10; ++i)
			if (sett>>i & 1) a[++cnt]=i;
		num[sett]=0;
		for (int i=cnt; i; --i) num[sett]=num[sett]*10+a[i];

		for (int i=1; i<=cnt; ++i)
		{
			flag[sett][a[i]]=true;
			for (int j=1; j<=cnt; ++j)
				if (i!=j)
				{
					flag[sett][a[i]*10+a[j]]=true;
					flag[sett][a[i]+a[j]]=true;
					for (int k=1; k<=cnt; ++k)
						if (i!=k && j!=k)
						{
							flag[sett][a[i]*10+a[j]+a[k]]=true;
							for (int p=1; p<=cnt; ++p)
								if (i!=p && j!=p && k!=p)
									flag[sett][a[i]*10+a[j]+a[k]*10+a[p]]=true;
						}
				}
		}
	}
	for (int i=0; i<1<<10; ++i) idx[i]=i;
	sort(idx, idx+(1<<10), cmp);
}
bool read()
{
	scanf("%d", &n);
	for (int i=1; i<=n; ++i) scanf("%d", &a[i]);
	return n;
}
void solve()
{
	for (int i=0; i<1<<10; ++i)
	{
		bool can=true;
		for (int j=1; j<=n && can; ++j) can&=flag[idx[i]][a[j]];
		if (can)
		{
			printf("%lld\n", num[idx[i]]);
			return;
		}
	}
}
int main()
{
	init();
	int casesum=0;
	while (read())
	{
		printf("Case %d: ", ++casesum);
		solve();
	}
	return 0;
}

C - Cards

题目描述:一副\(54\)张的扑克牌,随机排序,按顺序抽取,抽到大小王可以让他们变成任意一种花色(要当场决定),问期望抽多少张牌使得每种花色至少有\(A, B, C, D\)张。

solution
概率\(dp\),思路挺好想的,就记住每种花色剩下多少张,以及每个王变成了什么花色(或者还没抽到)

时间复杂度:\(O(13^4*5^2*4)\)(每次询问)

G - Pair of Touching Circles

题目描述:给定一个\(n \times m\)的网格图,在图中画两个圆,要求者两个圆的圆心要在格点上,半径要是整数(两个圆的半径不一定相等),整个圆要在网格图内,两个圆要相切,问方案数。

solution
先预处理出两个圆所形成的矩形的情况,以及每种情况对应的方案数,然后问题就变成了给定的网格图每种矩形有多少个。
预处理这部分可以枚举两个圆心横坐标的差以及纵坐标的差,判断是否能相切(其实就是判勾股数),然后再枚举其中一个圆的半径,就可以算出两个圆所对应的矩形,然后统计结果。

时间复杂度:\(O(能过)\)(因为勾股数不多)

H - Treasure Hunt

题目描述:在二维平面上,给出一个矩形的顶点坐标,在给定在矩形内的四个点,问这四个点要如何移动,使得移动后四个点的重心与矩形重心重合,并且移动的距离总和最小,求移动后的坐标。

solution
可以先把矩形的重心移到原点,然后算出四个点的重心,重心指向原点的向量就是四个点的移动的向量的总和,因为要移动的距离总和最小,因此所有点的移动的向量应与重心指向原点的向量平行,然后逐个移动即可。

时间复杂度:\(O(1)\)

#include <bits/stdc++.h>
using namespace std;

#define x first
#define y second
typedef long double LD;
const LD eps=1e-15;
const LD inf=1e18;

struct Point:public pair<LD, LD>
{

	Point(LD _x=0, LD _y=0):pair(_x, _y){}

	Point& operator += (Point c)
	{
		x+=c.x; y+=c.y;
		return *this;
	}

	Point& operator /= (LD c)
	{
		x/=c; y/=c;
		return *this;
	}

	Point& operator -= (Point c)
	{
		x-=c.x; y-=c.y;
		return *this;
	}

	Point operator + (Point c)
	{
		return Point(x+c.x, y+c.y);
	}

	Point operator - (Point c)
	{
		return Point(x-c.x, y-c.y);
	}

	Point operator * (LD c)
	{
		return Point(x*c, y*c);
	}

	Point operator / (LD c)
	{
		return Point(x/c, y/c);
	}

	bool zero()
	{
		return (fabs(x)<eps && fabs(x)<eps);
	}

	void rotate(LD angle)
	{
		LD tmp=x;
		x=-sin(angle)*y+cos(angle)*tmp;
		y=sin(angle)*tmp+cos(angle)*y;
	}
};

Point origin;
LD angle;
Point pos[10], rec[10], ans[10];

inline LD sqr(LD x)
{
	return x*x;
}
LD dis(Point b, Point c)
{
	return sqrt(sqr(b.x-c.x)+sqr(b.y-c.y));
}
bool read()
{
	for (int i=1; i<=4; ++i) scanf("%Lf%Lf", &pos[i].x, &pos[i].y);
	for (int i=1; i<=4; ++i) scanf("%Lf%Lf", &rec[i].x, &rec[i].y);
	for (int i=1; i<=4; ++i)
		if (fabs(pos[i].x)>eps || fabs(pos[i].y)>eps) return true;
	for (int i=1; i<=4; ++i)
		if (fabs(rec[i].x)>eps || fabs(rec[i].y)>eps) return true;
	return false;
}
void rotate()
{
	origin=Point(0, 0);
	for (int i=1; i<=4; ++i) origin+=rec[i];
	origin/=4;
	for (int i=1; i<=4; ++i)
	{
		rec[i]-=origin;
		pos[i]-=origin;
	}
	Point arrow=rec[2]-rec[1];
	angle=atan2(arrow.y, arrow.x);
	for (int i=1; i<=4; ++i)
	{
		rec[i].rotate(-angle);
		pos[i].rotate(-angle);
	}
}
void solve()
{
	rotate();
	Point total=Point(0, 0);
	Point boundx=Point(inf, -inf), boundy=Point(inf, -inf);
	for (int i=1; i<=4; ++i) total-=pos[i];
	for (int i=1; i<=4; ++i)
	{
		boundx.x=min(rec[i].x, boundx.x);
		boundx.y=max(rec[i].x, boundx.y);
		boundy.x=min(rec[i].y, boundy.x);
		boundy.y=max(rec[i].y, boundy.y);
	}

	LD answer=0;
	for (int i=1; i<=4; ++i)
	{
		Point bound=Point((total.x<0? boundx.x:boundx.y), (total.y<0? boundy.x:boundy.y));
		Point tmp=bound-pos[i];
		LD rate=min(LD(1.0), min(tmp.x/total.x, tmp.y/total.y));
		tmp=pos[i]+total*rate;
		total-=tmp-pos[i];
		answer+=dis(pos[i], tmp);
		pos[i]=tmp;
	}

	for (int i=1; i<=4; ++i)
	{
		pos[i].rotate(angle);
		pos[i]+=origin;
	}
	for (int i=1; i<=4; ++i) printf("%.12Lf %.12Lf\n", pos[i].x, pos[i].y);
	puts("");
}
void readans()
{
	for (int i=1; i<=4; ++i) scanf("%Lf%Lf", &ans[i].x, &ans[i].y);
	LD answer=0;
	for (int i=1; i<=4; ++i) answer+=dis(ans[i], pos[i]);
	printf("%.12Lf\n", answer);
}
int main()
{
	while (read()) solve();
	return 0;
}

I - Truchet Tiling

题目描述:有两种\(2 \times 2\)的地砖,如图,现在有\(n \times m\)块地砖拼在一起,以圆弧连成的线作为分界进行涂色,给出一些询问,每次给定一个坐标,问该坐标所在的区域的面积。

solution
将一块地砖分成三个区域,然后题目的意思做并查集,再处理出每个坐标所在的并查集。

时间复杂度:\(O(nm)\)

#include <bits/stdc++.h>
using namespace std;

const double PI=acos(-1);
const int maxn=110;

int n, m;
int dsu[maxn*maxn*4];
double area[maxn*maxn*4];
bool Map[maxn][maxn];
int pos[maxn*2][maxn*2];

void read()
{
	scanf("%d%d", &n, &m);
	for (int i=1; i<=n*m*3; i+=3) 
	{
		dsu[i]=dsu[i+1]=dsu[i+2]=-1;
		area[i]=area[i+2]=PI/4;
		area[i+1]=2*2-PI/2;
	}
	for (int i=1; i<=n; ++i)
		for (int j=1; j<=m; ++j)
			scanf("%1d", &Map[i][j]);
}
int dsu_find(int cur)
{
	return (dsu[cur]<0? cur:(dsu[cur]=dsu_find(dsu[cur])));
}
void merge(int u, int v)
{
	u=dsu_find(u);
	v=dsu_find(v);
	if (u==v) return;
	if (dsu[u]>dsu[v]) swap(u, v);
	dsu[u]+=dsu[v];
	dsu[v]=u;
	area[u]+=area[v];
}
void divide()
{
	for (int i=1; i<=n; ++i)
		for (int j=1; j<=m; ++j)
		{
			int cur=((i-1)*m+j-1)*3;
			if (i!=1)
			{
				int nx=((i-2)*m+j-1)*3;
				if (Map[i][j]==Map[i-1][j])
				{
					merge(nx+2, cur+1);
					merge(nx+3, cur+2);
				}
				else
				{
					merge(nx+3, cur+1);
					merge(nx+2, cur+2);
				}
			}
			if (j!=1)
			{
				int nx=((i-1)*m+j-2)*3;
				if (Map[i][j])
				{
					if (Map[i][j-1])
					{
						merge(nx+1, cur+2);
						merge(nx+2, cur+3);
					}
					else
					{
						merge(nx+2, cur+2);
						merge(nx+3, cur+3);
					}
				}
				else
				{
					if (Map[i][j-1])
					{
						merge(nx+1, cur+1);
						merge(nx+2, cur+2);
					}
					else
					{
						merge(nx+2, cur+1);
						merge(nx+3, cur+2);
					}
				}
			}
		}
}
void calc_pos()
{
	for (int i=1; i<=n; ++i)
		for (int j=1; j<=m; ++j)
		{
			pos[(i-1)*2][j*2-1]=pos[i*2-1][(j-1)*2]=pos[i*2-1][j*2]=pos[i*2][j*2-1]=0;
			if (Map[i][j])
			{
				pos[(i-1)*2][j*2]=((i-1)*m+j-1)*3+1;
				pos[(i-1)*2][(j-1)*2]=pos[i*2-1][j*2-1]=pos[i*2][j*2]=((i-1)*m+j-1)*3+2;
				pos[i*2][(j-1)*2]=((i-1)*m+j)*3;
			}
			else
			{
				pos[(i-1)*2][(j-1)*2]=((i-1)*m+j-1)*3+1;
				pos[(i-1)*2][j*2]=pos[i*2-1][j*2-1]=pos[i*2][(j-1)*2]=((i-1)*m+j-1)*3+2;
				pos[i*2][j*2]=((i-1)*m+j)*3;
			}
		}
		/*
	for (int i=0; i<=n*2; ++i, putchar('\n'))
		for (int j=0; j<=m*2; ++j)
			printf("%d ", pos[i][j]);
			*/
}
void solve()
{
	calc_pos();
	divide();
	int T;
	scanf("%d", &T);
	while (T--)
	{
		int x, y;
		scanf("%d%d", &x, &y);
		if (pos[x][y]==0) printf("%.4lf\n", 0.0);
		else printf("%.4lf\n", area[dsu_find(pos[x][y])]);
	}
}
int main()
{
	int casesum;
	scanf("%d", &casesum);
	for (int i=1; i<=casesum; ++i)
	{
		printf("Case %d:\n", i);
		read();
		solve();
	}
	return 0;
}

J - As Long as I Learn, I Live

solution
模拟。

时间复杂度:\(O(m)\)

posted @ 2018-11-01 23:00  GerynOhenz  阅读(66)  评论(0编辑  收藏