ZR1153

ZR1153

首先我们可以发现一个比较简单的容斥做法

直接暴力枚举\(2^m\)个限制强制不合法,算贡献

注意如果两个限制冲突那么答案为0

直接暴力差分就好了

这样就有了快乐的\(30\)分了

接下来考虑对容斥进行DP

把所有点区间按照右端点排序,如果出来两个颜色相同的区间一个包含了另外一个,那么大区间是没有用的,因为小区间满足条件大区间一定满足

我们设\(f_{i}\)表示满足第\(i\)个限制的带容斥系数的方案数

那么转移我们就枚举上一个没有交的区间

\[f_{i} =g_{;_i -1} + \sum_{co_i = co_j,l_j\ge r_i}f_j
\]

其中\(g_x\)表示\(1-x\)位置满足\(1-x\)的所有容斥之后的限制的前缀和

也就是说

\[g_i = g_{i - 1}\times s + \sum_{r_j = i}f_j
\]

就是看看第\(i\)位置上的限制满足还是不满足综合考虑的前缀和

继续回到求\(f_i\)的式子

既然\(i\)的这个限制要容斥,那么强制他不被满足,前面就是对所有和他没有交的限制求一个总的容斥

后面算有交的部分的贡献,必须满足和当前限制的颜色相同,

我们排序之后,有交的集合是一个区间,我们二分找到对应位置维护前缀和即可

#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<ctime>
#include<cmath>
#include<set>
#include<map>
#define LL long long
#define pii pair<int,int>
#define mk make_pair
#define fi first
#define se second
using namespace std;
const int N = 4e5 + 3;
const LL mod = 998244353;
struct seg{
	int li,ri;
	int xi;
}a[N],b[N];
vector <pii> co[N];
vector <LL> h[N];
LL f[N],g[N];
int n,m,s,cnt;
inline int read(){
	int v = 0,c = 1;char ch = getchar();
	while(!isdigit(ch)){
		if(ch == '-') c = -1;
		ch = getchar();
	}
	while(isdigit(ch)){
		v = v * 10 + ch - 48;
		ch = getchar();
	}
	return v * c;
}
inline bool cmp(seg x,seg y){
	return x.ri < y.ri || (x.ri == y.li && x.li > y.li);
}
inline LL find(int x,int rr){
//	printf("%d %d\n",x,rr);
	int l = 0,r = h[x].size() - 1,ans = -1;
	if(r < 0) return 0;
	while(l <= r){
		int mid = (l + r) >> 1;
		if(co[x][mid].se < rr) l = mid + 1,ans = mid;
		else r = mid - 1;
	}
//	printf("%d %d %d %lld\n",l,r,ans,ans == -1 ? h[x].back() : h[x].back() - h[x][ans]);
	return ans == -1 ? h[x].back() : h[x].back() - h[x][ans];
}
inline LL mo(LL x){
	if(x >= mod) x-= mod;
	return x;
}
int main(){
	n = read(),m = read(),s = read();
	for(int i = 1;i <= m;++i){
		a[i].li = read();
		a[i].ri = read();
		a[i].xi = read();
	}
	sort(a + 1,a + m + 1,cmp);
	for(int i = 1;i <= m;++i){
		bool flag = 0;
		if(!co[a[i].xi].size()) co[a[i].xi].push_back(mk(a[i].li,a[i].ri));
		else{
			pii x = co[a[i].xi].back();
			if(x.fi >= a[i].li && x.se <= a[i].ri) flag = 1;
			else co[a[i].xi].push_back(mk(a[i].li,a[i].ri));
		}
		if(!flag) b[++cnt] = a[i];
	}
	m = cnt;
	memcpy(a,b,sizeof(a));
//	puts("new::");
//	for(int i = 1;i <= m;++i) cerr << a[i].li << " " << a[i].ri << " " << a[i].xi << endl;
//	puts("next::");
	int now = 1;
	f[0] = g[0] = 1;
	for(int i = 1;i <= m;++i){
	//	cerr << "dsdas::"<< a[i].li << " " << a[i].ri << " " << a[i].xi << endl;
		for(;now < a[i].ri;now++) g[now] = (g[now] + g[now - 1] * s) % mod;
		f[i] = (-g[a[i].li - 1] + mod);
		f[i] -= find(a[i].xi,a[i].li);
		if(f[i] < 0) f[i] += mod;
		LL gg = h[a[i].xi].empty() ? 0 : h[a[i].xi].back();
		h[a[i].xi].push_back(mo(gg + f[i]));
		g[a[i].ri] = mo(g[a[i].ri] + f[i]);
	}
	for(;now <= n;++now) g[now] = (g[now] + g[now - 1] * s) % mod;
	printf("%lld\n",g[n]);
	return 0;
}