题解 [AGC018E] Sightseeing Plan

传送门

格路计数全家桶实锤了

官方题解
Luogu题解
上面题解中讲述了若干种常见格路计数方法，这里就不再写一遍了
本题需要注意的细节：
$len=x_2-x_1+y_2-y_1\color{red}{+1}$
枚举进入点和离开点因为是进入和离开所以要求到这个点的上/下一步在矩形外
复杂度 $O(V)$

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 2000010
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

namespace task{
	const ll mod=1e9+7;
	ll fac[N], inv[N], ans;
	int x1, x2, x3, x4, x5, x6;
	int y1, y2, y3, y4, y5, y6;
	inline ll C(int n, int k) {return fac[n]*inv[k]%mod*inv[n-k]%mod;}
	inline ll f(int x1, int y1, int x2, int y2) {return C(abs(y2-y1)+abs(x2-x1), abs(x2-x1));}
	inline ll upsqr(int x, int y, int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4) {
		return (f(x, y, x4+1, y4+1)-f(x, y, x2+1, y2)-f(x, y, x3, y3+1)+f(x, y, x1, y1))%mod;
	}
	inline ll downsqr(int x, int y, int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4) {
		return (f(x, y, x1-1, y1-1)-f(x, y, x2, y2-1)-f(x, y, x3-1, y3)+f(x, y, x4, y4))%mod;
	}
	void solve() {
		x1=read(); x2=read(); x3=read(); x4=read(); x5=read(); x6=read();
		y1=read(); y2=read(); y3=read(); y4=read(); y5=read(); y6=read();
		fac[0]=fac[1]=1; inv[0]=inv[1]=1;
		for (int i=2; i<N; ++i) fac[i]=fac[i-1]*i%mod;
		for (int i=2; i<N; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
		for (int i=2; i<N; ++i) inv[i]=inv[i-1]*inv[i]%mod;
		for (int i=x3; i<=x4; ++i) ans=(ans-(i+y3)*downsqr(i, y3-1, x1, y1, x2, y1, x1, y2, x2, y2)%mod*upsqr(i, y3, x5, y5, x6, y5, x5, y6, x6, y6))%mod;
		for (int i=y3; i<=y4; ++i) ans=(ans-(x3+i)*downsqr(x3-1, i, x1, y1, x2, y1, x1, y2, x2, y2)%mod*upsqr(x3, i, x5, y5, x6, y5, x5, y6, x6, y6))%mod;
		for (int i=x3; i<=x4; ++i) ans=(ans+(i+y4+1)*downsqr(i, y4, x1, y1, x2, y1, x1, y2, x2, y2)%mod*upsqr(i, y4+1, x5, y5, x6, y5, x5, y6, x6, y6))%mod;
		for (int i=y3; i<=y4; ++i) ans=(ans+(x4+i+1)*downsqr(x4, i, x1, y1, x2, y1, x1, y2, x2, y2)%mod*upsqr(x4+1, i, x5, y5, x6, y5, x5, y6, x6, y6))%mod;
		printf("%lld\n", (ans%mod+mod)%mod);
	}
}

signed main()
{
	task::solve();
	
	return 0;
}