FJNUOJ Yehan’s hole（容斥求路径数 + 逆元）题解

Description

Yehan is a angry grumpy rabbit, who likes jumping into the hole. This day,Yehan jumps again in the pit at home. Each time, he should jump from the hole at the coordinate (1,1) to (n,m), and he has to jump as the way : he can only jump (x, y) to (x+1, y) or (x, y+1). At his home, some holes are filled with water, he couldn’t jump in them. Yehan wants to know how many different way he could jump. If the paths of two ways are different, Yehan considers they are different.

Input

Multiple test dataThe first line of input contains two numbers n, m (2 <= n, m < 1e5)The second line of input contains a number q, the number of holes (q <= 15)The following q lines, the i-th line contains two numbers xi, yi (1 < x1 < x2 < x3 <……< xq < n) (1 < y1 < y2 < y3 < …… < yq < m)

Output

Output contains one line, the number of ways, because the result is too large, you should mod 1000000007

题意：从（1，1）走到（n，m），其中有些格子有水坑不能走（保证水坑坐标(1 < x1 < x2 < x3 <……< xq < n) (1 < y1 < y2 < y3 < …… < yq < m)，之前没看到这个坑死），问你有多少路径。

思路：对于x*y的方格，从左上走到右下的路径数为：C^x_x+y。这个高中组合应该讲过，因为横向必走x步，而且只要这x步确定了那么纵向怎么走必确定（可以自己试试），而横着走有x + y种可能。

所以这道题我们要做的是总路径减去走水坑的路径。走水坑路径和总路径做法一样，但要用容斥。因为这里要组合数取模1e+7，所以还要用到逆元。一开始很傻比直接算逆元，其实这里直接打表阶乘的逆元，具体看下面参考内容。

代码最后是数据。

参考：【逆元】

代码：

#include<cstdio>
#include<set>
#include<vector>
#include<cmath>
#include<queue>
#include<cstring>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = +;
const double INF = 0x3f3f3f3f;
const ll MOD = ;
struct node{
    int x, y;
}p[];
ll ans, n, m;
ll fac[maxn << ], inv[maxn << ];
int q;
int cmp(node a, node b){
    return a.x == b.x? a.y < b.y : a.x < b.x;
}
ll pmul(ll a, ll b){
    ll ans = ;
    while(b){
        if(b & ) ans = (ans * a) % MOD;
        a = a * a % MOD;
        b >>= ;
    }
    return ans;
}
void init(){    //阶乘的逆元
    fac[] = fac[] = ;
    for(ll i = ; i <= ; i++)
        fac[i] = fac[i - ] * i % MOD;
    inv[] = pmul(fac[], MOD - );
    for(ll i =  - ; i >= ; i--)
        inv[i] = inv[i + ] * (i + ) % MOD;
}
ll C(ll x, ll y){   //组合逆元
    return (fac[y] * inv[y - x] % MOD) * inv[x] % MOD;
}
void dfs(int id, int pre, ll temp, int times){
    ll tmp1, tmp2;
    times++;
    tmp1 = (temp * C(p[id].x - p[pre].x, p[id].x - p[pre].x + p[id].y - p[pre].y)) % MOD;
    tmp2 = (tmp1 * C(n - p[id].x, n - p[id].x + m - p[id].y)) % MOD;
    if(times & ) ans = (ans + tmp2) % MOD;
    else ans = (ans - tmp2) % MOD;
    for(int i = id + ; i <= q; i++){
        dfs(i, id, tmp1, times);
    }
}
int main(){
    init();
    while(~scanf("%lld%lld", &n, &m)){
        scanf("%d", &q);
        p[].x = , p[].y = ;
        for(int i = ; i <= q; i++){
            scanf("%d%d", &p[i].x, &p[i].y);
        }
        ans = ;
        for(int i = ; i <= q; i++){
            dfs(i, , , );    //now pre mul time
        }
        printf("%lld\n", ((C(n - , n + m - ) - ans) % MOD + MOD) % MOD);
    }
    return ;
}

/*
input
6122 61753
2
1957 18165
4448 30108
91557 89797
9
11169 5062
30315 34046
37127 36827
44369 50521
62770 58297
64008 62857
79378 67405
88132 86222
89483 87524
74355 52758
12
1193 7922
11316 9175
19053 11037
38189 11344
38317 15083
40095 19756
41161 24874
47120 30594
50188 34496
51327 36374
53291 50539
66554 51069
56766 19695
7
2012 1210
12159 11135
18759 13249
40035 13828
41494 13959
45882 18241
49535 19379
31178 59737
7
698 14597
7306 14856
8198 19316
9000 45862
10879 53006
12002 54423
24634 58820
81872 8035
4
32720 1451
40464 4253
51261 6151
72014 6683
21393 42250
13
422 1192
478 2683
3265 3684
4422 4366
4760 13041
6586 22349
6803 24676
7273 25162
8875 29191
13875 32524
14791 33611
17168 34162
21102 39838
5572 26973
9
1760 697
2106 4583
2141 6788
2438 6790
3811 7301
4293 11943
4607 15554
5164 15929
5529 18282

output
390661224
872660150
97529549
252410050
742624594
697140589
674450439
109291758
*/