CodeForces - 559C Gerald and Giant Chess

最新推荐文章于 2022-02-13 18:03:46 发布

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本文探讨了一种名为巨棋的游戏，在此游戏中玩家需避开黑色障碍单元格，将棋子从棋盘左上角移至右下角。文章提供了一个算法解决方案，利用动态规划和组合数学来计算所有可能的走法数量。

Giant chess is quite common in Geraldion. We will not delve into the rules of the game, we'll just say that the game takes place on an h × w field, and it is painted in two colors, but not like in chess. Almost all cells of the field are white and only some of them are black. Currently Gerald is finishing a game of giant chess against his friend Pollard. Gerald has almost won, and the only thing he needs to win is to bring the pawn from the upper left corner of the board, where it is now standing, to the lower right corner. Gerald is so confident of victory that he became interested, in how many ways can he win?

The pawn, which Gerald has got left can go in two ways: one cell down or one cell to the right. In addition, it can not go to the black cells, otherwise the Gerald still loses. There are no other pawns or pieces left on the field, so that, according to the rules of giant chess Gerald moves his pawn until the game is over, and Pollard is just watching this process.

Input

The first line of the input contains three integers: h, w, n — the sides of the board and the number of black cells (1 ≤ h, w ≤ 10⁵, 1 ≤ n ≤ 2000).

Next n lines contain the description of black cells. The i-th of these lines contains numbers r_i, c_i (1 ≤ r_i ≤ h, 1 ≤ c_i ≤ w) — the number of the row and column of the i-th cell.

It is guaranteed that the upper left and lower right cell are white and all cells in the description are distinct.

Output

Print a single line — the remainder of the number of ways to move Gerald's pawn from the upper left to the lower right corner modulo 10⁹ + 7.

Example

Input

3 4 2
2 2
2 3

Output

Input

Output

545732279

首先，一定是dp。

这道题很像过河卒，但是空间超了。

如果没有坏点，从（1,1）走到（n，m）有C(n-1,m+n-2)种方法。

设f[i]表示从1,1到第i个坏点的方案数。

f[i]=C(x[i]-1,y[i]+x[i]-2)-f[j]*C(x[i]-x[j],x[i]+y[i]-x[j]-y[j]).

当然，需要先排序。

#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;
const long long mod=1e9+7;
const int N=2005;
int n,m,t;
long long f[N],jc[200005];
struct node
{
	long long x,y;
}a[N];
bool cmp(node d,node e)
{
	if(d.x==e.x)
		return d.y<e.y;
	return d.x<e.x;
}
long long ksm(long long x,long long y)
{
	long long res=1;
	while(y)
	{
		if(y&1)
			res*=x,res%=mod;
		y>>=1;
		x*=x,x%=mod;
	}
	return res;
}
long long c(long long x,long long y)
{
	if(x==0)
		return 1;
	return jc[y]*ksm((jc[x]*jc[y-x])%mod,mod-2)%mod;
}
int main()
{
	scanf("%d%d%d",&n,&m,&t);
	for(int i=1;i<=t;i++)
		scanf("%I64d%I64d",&a[i].x,&a[i].y);
	sort(a+1,a+t+1,cmp);
	a[0].x=a[0].y=1;
	a[t+1].x=n,a[t+1].y=m;
	jc[0]=1;
	for(int i=1;i<=n+m;i++)
		jc[i]=(jc[i-1]*i)%mod;
	for(int i=1;i<=t+1;i++)
	{
		f[i]=c(a[i].x-a[0].x,a[i].x+a[i].y-a[0].x-a[0].y);
		for(int j=1;j<i;j++)
			if(a[j].y<=a[i].y)
				f[i]-=f[j]*c(a[i].x-a[j].x,a[i].x+a[i].y-a[j].x-a[j].y)%mod,f[i]=(f[i]%mod+mod)%mod;
	}
	printf("%I64d\n",f[t+1]);
	return 0;
}