CF 560e Gerald and Giant Chess
题意:在h×w的棋盘中从左上角走到右下角,只能向右或向下走,有n个点不可以经过,一共有多少种方案。
解法:dp。先对点按横坐标排序(横坐标相等按纵坐标,也可以反过来)dp[i]表示不经过其他非法点走到第i个非法点的路径数,则有dp方程:dp[i] = c(point[i].x + point[i].y - 2, point[i].x - 1) - Σj = 0...i - 1 dp[j] * c(point[i].x - point[j].x + point[i].y - point[j].y, point[i].x - point[j].x),c表示组合数,c(point[i].x + point[i].y - 2, point[i].x - 1)表示从起点到该非法点i的所有路径,再减去在该点之前的每个点j,从起点到点j的路径数乘以从点j到点i的路径数,即为所求,将终点加入点集中,则dp[n]为答案。
而对于组合数的求法,这道题无法用杨辉三角打表,所以用阶乘来求,即c(n, m) = n! / (m! * (n - m)!),但由于有取模运算,不能直接做除法,所以转化为乘以分母的逆元,m = n mod p的逆元为mp-2。
代码:
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define LL long long
using namespace std;
struct node
{
LL x, y;
bool operator < (const node &tmp) const
{
if(x == tmp.x)
return y < tmp.y;
return x < tmp.x;
}
}point[2005];
const LL mod = 1e9 + 7;
LL jc[200005];//阶乘
void init()
{
jc[0] = 1;
for(int i = 1; i < 200005; i++)
{
jc[i] = (jc[i - 1] * i) % mod;
}
}
LL Pow(LL n)//求逆元
{
n %= mod;
LL x = 1e9 + 5;
LL res = 1;
while(x)
{
if(x & 1)
res = res * n % mod;
n = n * n % mod;
x >>= 1;
}
return res;
}
LL c(int x, int y)
{
if(x < 0)
return 0;
if(y < 0)
return 0;
if(y > x)
return 0;
//不判断以上三个条件会RE
return jc[x] * Pow(jc[y] * jc[x - y] % mod) % mod;
}
LL dp[2005];
int main()
{
init();
int h, w, n;
while(~scanf("%d%d%d", &h, &w, &n))
{
memset(dp, 0, sizeof dp);
for(int i = 0; i < n; i++)
{
cin >> point[i].x >> point[i].y;
}
point[n].x = h, point[n].y = w;
sort(point, point + n);
LL ans = 0;
for(int i = 0; i <= n; i++)
{
LL tmp = c(point[i].x + point[i].y - 2, point[i].x - 1);
for(int j = 0; j < i; j++)
{
tmp += mod - (dp[j] * c(point[i].x - point[j].x + point[i].y - point[j].y, point[i].x - point[j].x) % mod);
tmp %= mod;
}
dp[i] = tmp;
}
cout << dp[n] << endl;
}
return 0;
}
CF不能交lld真蛋疼……
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