Codeforces Round #313 (Div. 1) C. Gerald and Giant Chess DP

C. Gerald and Giant Chess

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/559/problem/C

Description

Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to . Then he measured the length of its sides, and found that each of them is equal to an integer number of centimeters. There the properties of the hexagon ended and Gerald decided to draw on it.

He painted a few lines, parallel to the sides of the hexagon. The lines split the hexagon into regular triangles with sides of 1 centimeter. Now Gerald wonders how many triangles he has got. But there were so many of them that Gerald lost the track of his counting. Help the boy count the triangles.

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 ri, ci (1 ≤ ri ≤ h, 1 ≤ ci ≤ 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 modulo109 + 7.

Sample Input

3 4 2
2 2
2 3

Sample Output

2

HINT

题意

一个n*m的图，让你从左上角走到右下角，有一些点不能经过，问你有多少种方法

题解：

dp[i]表示不经过黑点的方案数

fi=Cxixi+yi−∑(xj<=xi,yj<=yi)C(xi−xj)(xi−xj)+(yi−yj)∗fj

然后一路小跑就好了

http://blog.csdn.net/popoqqq/article/details/46121519

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 1010000
#define M 10000
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//**************************************************************************************

struct Point
{
    long long x,y;
}points[];
bool cmp(Point a,Point b)
{
    if(a.x==b.x)
        return a.y<b.y;
    return a.x<b.x;
}
ll p=mod;
ll fac[maxn];
ll qpow(ll a,ll b)
{
    ll ans=;a%=mod;
    for(ll i=b;i;i>>=,a=a*a%mod)
        if(i&)ans=ans*a%mod;
    return ans;
}
ll C(ll n,ll m)
{
    if(m>n||m<)return ;
    ll s1=fac[n],s2=fac[n-m]*fac[m]%mod;
    return s1*qpow(s2,mod-)%mod;
}
ll f[maxn];
int main()
{
    fac[]=;
    for(int i=;i<maxn;i++)
        fac[i]=fac[i-]*i%mod;
    int n=read(),m=read(),k=read();
    for(int i=;i<=k;i++)
    {
        points[i].x=read();
        points[i].y=read();
        points[i].x-=;
        points[i].y-=;
    }
    points[++k].x=n-;
    points[k].y=m-;
    sort(points+,points+k+,cmp);
    for(int i=;i<=k;i++)
    {
        f[i]=C(points[i].x+points[i].y,points[i].x);
        for(int j=;j<i;j++)
        {
            if(points[j].y<=points[i].y)
            {
                f[i]+=(p-f[j]*C(points[i].x-points[j].x+points[i].y-points[j].y,points[i].x-points[j].x)%p);
                f[i]%=p;
            }
        }
    }
    cout<<f[k]%p<<endl;
}