【题解】CF559C C. Gerald and Giant Chess(容斥+格路问题)

【题解】CF559C C. Gerald and Giant Chess(容斥+格路问题)

55336399 Practice: Winlere 559C - 22 GNU C++11 Accepted 186 ms 1608 KB 2019-06-09 17:03:21 2019-06-09 17:03:21

$dp(i,j)$表示从原点到此有多少条合法的路径。转移从所有可以阻挡住他的点来。

//@winlere
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define x first
#define y second

using namespace std;  typedef long long ll;
inline int qr(){
register int ret=0,f=0;
register char c=getchar();
while(c<48||c>57)f|=c==45,c=getchar();
while(c>=48&&c<=57) ret=ret*10+c-48,c=getchar();
return f?-ret:ret;
}
const int maxn=2e5+5;
const int mod=1e9+7;
int fac[maxn],inv[maxn],dp[2001],n;
pair < int , int > data[2001];
inline int ksm(int base,const int&p){register int ret=1;
for(register int t=p;t;t>>=1,base=1ll*base*base%mod)if(t&1) ret=1ll*ret*base%mod;
return ret;
}

inline int c(const int&n,const int&m){return n<m?0:1ll*fac[n]*inv[m]%mod*1ll*inv[n-m]%mod;}
inline int grid(const int&x,const int&y){return c(x+y,x);}
int main(){
fac[0]=inv[0]=1;
for(register int t=1;t<maxn;++t) fac[t]=1ll*fac[t-1]*t%mod,inv[t]=ksm(fac[t],mod-2);

data[0].x=qr();data[0].y=qr();n=qr();
for(register int t=1;t<=n;++t) data[t].x=qr(),data[t].y=qr();

sort(data,data+n+1);

for(register int t=0;t<=n+1;++t){
dp[t]=grid(data[t].x-1,data[t].y-1);
for(register int i=0;i<t;++i){
dp[t]=(dp[t]-1ll*dp[i]*grid(data[t].x-data[i].x,data[t].y-data[i].y)%mod)%mod;
if(dp[t]<0) dp[t]+=mod;
}
}
cout<<dp[n]<<endl;

return 0;
}


posted @ 2019-06-09 22:09  谁是鸽王  阅读(142)  评论(0编辑  收藏