题意

题目链接

分析

• 又是一个二分图匹配的问题，考虑霍尔定理。
• 根据套路我们知道只需要检查 "区间的并是一段连续的区间" 这些子集。
• 首先将环倍长。考虑枚举答案的区间并的右端点 \(r\)，显然 \(r\) 应该在某个区间的右端点上。我们想要判断是否存在一个 \(l\) 使得 \(r-l+1\le m\) 且 \(\sum\limits_{l\le L_i,R_i\le r}a_i>r-l+1\) ，扫描线+线段树 即可。

• 有一类特殊情况：区间的并是整个环，这时它在序列上的表示长度可能不是这个并的真实长度(因为可能会有同一个区间出现两次)，我们此时只需要特判 \(\sum a>m\) 即可。

• 总时间复杂度为 \(O(nlogn)\)。

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define go(u) for(int i = head[u], v = e[i].to; i; i=e[i].lst, v=e[i].to)
#define rep(i, a, b) for(int i = a; i <= b; ++i)
#define pb push_back
#define re(x) memset(x, 0, sizeof x)
inline int gi() {int x = 0,f = 1;char ch = getchar();while(!isdigit(ch)) { if(ch == '-') f = -1; ch = getchar();}while(isdigit(ch)) { x = (x << 3) + (x << 1) + ch - 48; ch = getchar();}return x * f;
}
template <typename T> inline void Max(T &a, T b){if(a < b) a = b;}
template <typename T> inline void Min(T &a, T b){if(a > b) a = b;}
const int N = 4e5 + 7;
int T, n, m, vc;
LL adv[N << 2], mx[N << 2], V[N << 1];
#define Ls o << 1
#define Rs (o << 1 | 1)
void st1(int o, LL v) {adv[o] += v;mx[o] += v;
}
void pushdown(int o) {if(!adv[o]) return;st1(Ls, adv[o]);st1(Rs, adv[o]);adv[o] = 0;
}
void pushup(int o) {mx[o] = max(mx[Ls], mx[Rs]);
}
void build(int l, int r,int o){adv[o] = 0;if(l == r) {mx[o] = V[l];return;}int mid = l + r >> 1;build(l, mid, Ls);build(mid + 1, r, Rs);pushup(o);
}
void modify(int L, int R, int l, int r,int o, LL v) {if(L > R) return;if(L <= l && r <= R) {st1(o, v);return;}pushdown(o);int mid = l + r >> 1;if(L <= mid) modify(L, R, l, mid, Ls, v);if(R > mid)  modify(L, R, mid + 1, r, Rs, v);pushup(o);
}
int query(int L, int R, int l, int r, int o) {if(L > R) return 0;if(L <= l && r <= R) return mx[o];pushdown(o);int mid = l + r >> 1;if(R <= mid) return query(L, R, l, mid, Ls);if(L > mid)  return query(L, R, mid + 1, r, Rs);return max(query(L, R, l, mid, Ls), query(L, R, mid + 1, r, Rs));
}
struct qs {int l, r;LL a;bool operator <(const qs &rhs) const {return r < rhs.r;}
}q[N];
int main() {T = gi();while(T--) {n = gi(), m = gi();int ndc = n, sum = 0;vc = 0;rep(i, 1, n) {q[i].l = gi(), q[i].r = gi(), q[i].a = gi();sum += q[i].a;if(q[i].l <= q[i].r) q[++ndc] = (qs){ q[i].l + m, q[i].r + m, q[i].a};else q[i].r += m;}n = ndc;if(sum > m) { puts("No"); goto A;}sort(q + 1, q + 1 + n);rep(i, 1, n) V[++vc] = q[i].l, V[++vc] = q[i].r;sort(V + 1, V + 1 + vc);vc = unique(V + 1, V + 1 + vc) - V - 1;build(1, vc, 1);rep(i, 1, n) {q[i].l = lower_bound(V + 1, V + 1 + vc, q[i].l) - V;int pos = lower_bound(V + 1, V + 1 + vc, q[i].r - m + 1) - V;q[i].r = lower_bound(V + 1, V + 1 + vc, q[i].r) - V;modify(1, q[i].l, 1, vc, 1, q[i].a);LL res = query(pos, q[i].r, 1, vc, 1);if(res > V[q[i].r] + 1){ puts("No"); goto A;}}puts("Yes");A:;}return 0;
}