牛客 多校第一场 A Monotonic Matrix

题目描述
Count the number of n x m matrices A satisfying the following condition modulo (109+7).
* Ai, j ∈ {0, 1, 2} for all 1 ≤ i ≤ n, 1 ≤ j ≤ m.
* Ai, j ≤ Ai + 1, j for all 1 ≤ i < n, 1 ≤ j ≤ m.
* Ai, j ≤ Ai, j + 1 for all 1 ≤ i ≤ n, 1 ≤ j < m.
输入描述:
The input consists of several test cases and is terminated by end-of-file.
Each test case contains two integers n and m.
输出描述:
For each test case, print an integer which denotes the result.
示例1
输入

复制
1 2
2 2
1000 1000
输出

复制
6
20
540949876
备注:
* 1 ≤ n, m ≤ 103
* The number of test cases does not exceed 105.

题意

给你一个nxm的矩阵让你向其中填{0,1,2}三个数且满足Ai,j⩽Ai+1,j

，Ai,j⩽Ai,j+1有几种填法。

从起点a1（0，0） 终点b1（n, m）画一条非降路径，然后0，1是沿着网格走的过程。

从起点a2（-1，1） 终点b2（n-1, m+1）画一条非降路径，然后1，2是沿着网格走的过程。

Lindström–Gessel–Viennot引理我们就可以求出2条严格不相交的路径的方案数

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;     
	
typedef long long ll;
const int maxn = 2e3+5;
const ll mod = 1e9+7;
ll c[maxn][maxn];
     
void init()
{
    c[0][0] = 1;
    for(int i = 1; i < maxn; i++)
    {
        c[i][0] = 1;
        for(int j = 1; j <= i; j++)
            c[i][j] = (c[i-1][j]+c[i-1][j-1])%mod;
    }
}
     
int main()
{
    init();
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        ll  ans;
        ans = (c[n+m][n]*c[n+m][n]%mod-c[n+m][n+1]*c[n+m][m+1]%mod+mod)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}