Codeforces Round #313 (Div. 2) E. Gerald and Giant Chess (Lucas + dp)

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

 //#pragma comment(linker, "/STACK:102400000, 102400000")
 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 typedef __int64 LL;
 typedef pair <int, int> P;
 const int N = 2e3 + ;
 struct Node {
     LL x, y;
     bool operator <(const Node &cmp) const {
         return x == cmp.x ? y < cmp.y : x < cmp.x;
     }
 }node[N];
 LL mod = 1e9 + ;
 LL dp[N]; //经过i点不经过其他点的case数
 LL f[]; //阶乘
 
 LL Pow(LL a , LL n , LL mod) {
     LL res = ;
     while(n) {
         if(n & )
             res = res * a % mod;
         a = a * a % mod;
         n >>= ;
     }
     return res;
 }
 
 LL Comb(LL a , LL b , LL mod) {
     if(a < b) {
         return ;
     }
     if(a == b) {
         return ;
     }
     return f[a] * Pow(f[a - b] * f[b] % mod, mod - , mod) % mod;
 }
 
 LL Lucas(LL n , LL m , LL mod) {
     LL ans = ;
     while(m && n && ans) {
         ans = (ans * Comb(n % mod , m % mod , mod)) % mod;
         n /= mod;
         m /= mod;
     }
     return ans;
 }
 
 int main()
 {
     f[] = ;
     for(LL i = ; i <= ; ++i) {
         f[i] = f[i - ] * i % mod;
     }
     LL row, col, sum;
     int n;
     scanf("%lld %lld %d", &row, &col, &n);
     for(int i = ; i <= n; ++i) {
         scanf("%lld %lld", &node[i].x, &node[i].y);
         node[i].x--, node[i].y--;
     }
     sort(node + , node + n + );
     for(int i = ; i <= n; ++i) {
         sum = ;
         for(int j = ; j < i; ++j) {
             if(node[i].x >= node[j].x && node[i].y >= node[j].y) {
                 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;
             }
         }
         dp[i] = ((Lucas(node[i].x + node[i].y, node[i].x, mod) - sum) % mod + mod) % mod;
     }
     sum = ;
     for(int i = ; i <= n; ++i) {
         sum = (dp[i]*Lucas(row + col -  - node[i].x - node[i].y, row -  - node[i].x, mod) % mod + sum) % mod;
     }
     printf("%lld\n", ((Lucas(row + col - , col - , mod) - sum) % mod + mod) % mod);
     return ;
 }