【UOJ #209】【UER #6】票数统计

做比赛的时候完全没想到怎么处理同时有前缀和后缀的限制。。智商不够啊QAQ。。其实根据数据范围就能猜出来，因为直接做的复杂度O(n)很明显n不可能给5000，所以应该枚举一下总人数，这样对后缀的限制就转化为对前缀的限制了。。然后x=y的限制取最大那一个，然后简单容斥一下即可。。这个题确实妙啊，几个转弯都很巧妙。。

这个题复杂度应该写成Tmn的，然后我偷懒写了个桶，就是Tn^2的，常数大了一点。。

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define erp(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
const int mo = 998244353;
const int MAXN = 5005;
int cas;
int n, m, m1, m2, mx, M;

int jc[MAXN], rjc[MAXN];
void makeit()
{
jc[0]=jc[1]=rjc[0]=rjc[1]=1;
rep(i,2,5000)jc[i]=1ll*jc[i-1]*i%mo, rjc[i]=1ll*(mo-mo/i)*rjc[mo%i]%mo;
rep(i,2,5000)rjc[i]=1ll*rjc[i]*rjc[i-1]%mo;
}
inline int C(int n,int r)
{
if(r<0||r>n) return 0;
return 1ll*jc
*rjc[r]%mo*rjc[n-r]%mo;
}

struct Lim {
int x, y;
bool operator < (const Lim&b) const { return x < b.x; }
} L[MAXN], R[MAXN], R1[MAXN], A[MAXN];

int h[MAXN];
bool setlim(int i,int x) { if(i<0||x<0) return 0; if(~h[i]&&h[i]!=x) return 0; return h[i]=x, 1; }

int calc(int num)
{
memset(h,-1,sizeof h);
setlim(0,0);
setlim(n,num);
rep(i,1,m1) if (!setlim(L[i].x, L[i].y)) return 0;
rep(i,1,m2) if (!setlim(R1[i].x, R1[i].y)) return 0;
int pre = 0, id = 0, ans = 1;
rep(i,1,n) if (~h[i]) { if(h[i]<pre)return 0; pre = h[i]; }
rep(i,1,n)
{
if (h[i]==-1) continue;
ans = 1ll*ans*C(i-id, h[i]-h[id])%mo;
id = i;
}
return ans;
}

int solve(int num)
{
int tot = num;
rep(i,1,m2) R1[i] = (Lim){n-R[i].x, tot-R[i].y};
if (!mx) return calc(num);
L[++m1] = (Lim){mx, mx};
int ans = calc(num);
L[m1] = (Lim){n-mx, tot-mx};
(ans += calc(num)) %= mo;
L[++m1] = (Lim){mx, mx};
(ans += mo-calc(num)) %= mo;
m1 -= 2;
return ans;
}

bool checkbad()
{
rep(i,2,m1) if(L[i].y<L[i-1].y) return 1;
rep(i,2,m2) if(R[i].y<R[i-1].y) return 1;
return 0;
}

void work()
{
scanf("%d%d",&n,&m);
m1 = m2 = mx = 0;
rep(i,1,m)
{
int x,y;
scanf("%d%d",&x,&y);
if (x>y) L[++m1] = (Lim){x,y};
else if (y>x) R[++m2] = (Lim){y,x};
else mx = max(mx, x);
}
sort(L+1, L+m1+1);
sort(R+1, R+m2+1);
if (checkbad()) { puts("0"); return; }
int ans = 0;
rep(i,mx,n)
(ans += solve(i)) %= mo;
printf("%d\n", ans);
}

int main()
{
makeit();
scanf("%d",&cas);
while (cas--)
work();
return 0;
}