SRM614 Div1 Hard

SRM614 Div1 Hard

题目描述

Solution

显然有：
$E(x,y)=(E(x-1,y)+E(x,y-1))/2+1$
直接高斯消元时间复杂度 $O((nm)^3)$。

可以发现这种做法十分浪费，消元之后会有大量冗余元素，即零行，我们考虑消去这些不必要状态。

我们发现可以只保留最后一行和最后一列的未知数，用上面的式子表示其他没有保留的格子，再计算答案即可。

时间复杂度 $O((n+m)^3)$。

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>

#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se second

using namespace std;

template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }

typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;

const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int MAXN=205;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{
	int f=1,x=0; char c=getchar();
	while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }
	while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }
	return x*f;
}
class TorusSailing
{
	lod a[MAXN][MAXN],f[MAXN][MAXN][MAXN];
	private:
		bool solve(int n)
		{
			for (int i=0;i<n;i++)
			{
				int maxj=i;
				for (int j=i+1;j<n;j++)
				    if (fabs(a[j][i])-fabs(a[maxj][i])>eps) maxj=j;
				if (fabs(a[maxj][i])<eps) return 0;
				if (maxj!=i)
					for (int j=0;j<=n;j++) swap(a[maxj][j],a[i][j]);
				for (int j=i+1;j<n;j++)
				{
					double t=a[j][i]/a[i][i];
					for (int k=i;k<=n;k++) a[j][k]-=t*a[i][k];
				}
			}
			for (int i=n-1;i>=0;i--)
			{
				for (int j=i+1;j<n;j++) a[i][n]-=a[j][n]*a[i][j];
				a[i][n]/=a[i][i];
			}
			return 1;
		} 
		int upd(int x,int y,int mods){ return x+y<0?x+y+mods:x+y; }
	public:
		lod expectedTime(int n,int m,int X,int Y)
		{
			n--,m--;
			for (int i=0;i<n;i++) f[i][m][i]=1;
			for (int i=0;i<=m;i++) f[n][i][i+n]=1;
			
			for (int i=0;i<=n;i++)
				for (int j=0;j<=m;j++)
				{
					if (!i&&!j) continue;
					for (int k=0;k<=n+m+1;k++)
						f[i][j][k]=(f[upd(i,-1,n+1)][j][k]+f[i][upd(j,-1,m+1)][k])*0.5;
					f[i][j][n+m+1]++;					
//					for (int k=0;k<=n+m+1;k++) cout<<setw(6)<<f[i][j][k];
//					cout<<endl;
				} 
			
			for (int i=0;i<n;i++)
			{
				for (int j=0;j<=n+m+1;j++) a[i][j]=f[i][m][j];
				a[i][n+m+1]=-a[i][n+m+1];
				a[i][i]--;
			}			
			for (int i=0;i<=m;i++)
			{
				for (int j=0;j<=n+m+1;j++) a[i+n][j]=f[n][i][j];
				a[i+n][n+m+1]=-a[i+n][n+m+1];
				a[i+n][i+n]--;
			}			
			/*
			for (int i=0;i<=n+m;i++)
			{
				for (int j=0;j<=n+m+1;j++) cout<<setw(6)<<a[i][j];
				cout<<endl;
			}
			*/
			solve(n+m+1);
			lod ans=f[X][Y][n+m+1];
			for (int i=0;i<=n+m;i++) ans+=f[X][Y][i]*a[i][n+m+1];
			return ans;
		}
};
/*
TorusSailing solve;
int main()
{
	int n=read(),m=read(),X=read(),Y=read();
	printf("%.11Lf\n",solve.expectedTime(n,m,X,Y));
	return 0;
}
*/