题目链接：http://codeforces.com/contest/560/problem/E
给你一个n*m的网格，有k个坏点，问你从(1,1)到(n,m)不经过坏点有多少条路径。
先把这些坏点排序一下。
dp[i]表示从(1,1)到第i个坏点且不经过其他坏点的路径数目。
dp[i] = Lucas(x[i], y[i]) - sum(dp[j]*Lucas(x[i]-x[j], y[i]-x[j])) , x[j] <= x[i] && y[j] <= y[i] //到i点所有的路径数目 - 经过其他点的路径数目
1 //#pragma comment(linker, "/STACK:102400000, 102400000") 2 #include <algorithm> 3 #include <iostream> 4 #include <cstdlib> 5 #include <cstring> 6 #include <cstdio> 7 #include <vector> 8 #include <cmath> 9 #include <ctime> 10 #include <list> 11 #include <set> 12 #include <map> 13 using namespace std; 14 typedef __int64 LL; 15 typedef pair <int, int> P; 16 const int N = 2e3 + 5; 17 struct Node { 18 LL x, y; 19 bool operator <(const Node &cmp) const { 20 return x == cmp.x ? y < cmp.y : x < cmp.x; 21 } 22 }node[N]; 23 LL mod = 1e9 + 7; 24 LL dp[N]; //经过i点不经过其他点的case数 25 LL f[200005]; //阶乘 26 27 LL Pow(LL a , LL n , LL mod) { 28 LL res = 1; 29 while(n) { 30 if(n & 1) 31 res = res * a % mod; 32 a = a * a % mod; 33 n >>= 1; 34 } 35 return res; 36 } 37 38 LL Comb(LL a , LL b , LL mod) { 39 if(a < b) { 40 return 0; 41 } 42 if(a == b) { 43 return 1; 44 } 45 return f[a] * Pow(f[a - b] * f[b] % mod, mod - 2, mod) % mod; 46 } 47 48 LL Lucas(LL n , LL m , LL mod) { 49 LL ans = 1; 50 while(m && n && ans) { 51 ans = (ans * Comb(n % mod , m % mod , mod)) % mod; 52 n /= mod; 53 m /= mod; 54 } 55 return ans; 56 } 57 58 int main() 59 { 60 f[0] = 1; 61 for(LL i = 1; i <= 200000; ++i) { 62 f[i] = f[i - 1] * i % mod; 63 } 64 LL row, col, sum; 65 int n; 66 scanf("%lld %lld %d", &row, &col, &n); 67 for(int i = 1; i <= n; ++i) { 68 scanf("%lld %lld", &node[i].x, &node[i].y); 69 node[i].x--, node[i].y--; 70 } 71 sort(node + 1, node + n + 1); 72 for(int i = 1; i <= n; ++i) { 73 sum = 0; 74 for(int j = 1; j < i; ++j) { 75 if(node[i].x >= node[j].x && node[i].y >= node[j].y) { 76 sum = (dp[j]*Lucas(node[i].x + node[i].y - node[j].x - node[j].y, node[i].x - node[j].x, mod) % mod + sum) % mod; 77 } 78 } 79 dp[i] = ((Lucas(node[i].x + node[i].y, node[i].x, mod) - sum) % mod + mod) % mod; 80 } 81 sum = 0; 82 for(int i = 1; i <= n; ++i) { 83 sum = (dp[i]*Lucas(row + col - 2 - node[i].x - node[i].y, row - 1 - node[i].x, mod) % mod + sum) % mod; 84 } 85 printf("%lld ", ((Lucas(row + col - 2, col - 1, mod) - sum) % mod + mod) % mod); 86 return 0; 87 }