[CF482B]Interesting Array

题目大意：构造一个序列$S$，有$m$条限制，每条为$l\;r\;q$，表示$\&_{i=l}^r S_i=q$

题解：每条限制就把$[l,r]$内的数或上$q$，最后判断就行了

卡点：我又写了一课$O(n^2\log_2n)$的线段树。。。

C++ Code：

#include <cstdio>
#include <cstdlib>
#define maxn 100010
const int inf = (1 << 30) - 1;
namespace SgT {int n;int L, R, q;int V[maxn << 2], tg[maxn << 2];inline void pushdown(int rt) {int &x = tg[rt];V[rt << 1] |= x;V[rt << 1 | 1] |= x;tg[rt << 1] |= x;tg[rt << 1 | 1] |= x;x = 0;}void modify(int rt, int l, int r) {if (L <= l && R >= r) {V[rt] |= q;tg[rt] |= q;return ;}if (tg[rt]) pushdown(rt);int mid = l + r >> 1;if (L <= mid) modify(rt << 1, l, mid);if (R > mid) modify(rt << 1 | 1, mid + 1, r);V[rt] = V[rt << 1] & V[rt << 1 | 1];}inline void add(int ll, int rr, int qq) {L = ll, R = rr, q = qq;modify(1, 1, n);}int query(int rt, int l, int r) {if (L <= l && R >= r) return V[rt];if (tg[rt]) pushdown(rt);int mid = l + r >> 1, ans = inf;if (L <= mid) ans = query(rt << 1, l, mid);if (R > mid) ans &= query(rt << 1 | 1, mid + 1, r);return ans;}inline int ask(int ll, int rr) {L = ll, R = rr;return query(1, 1, n);}
}
using SgT::add;
using SgT::ask;
int n, m;
int l[maxn], r[maxn], q[maxn];
int main() {scanf("%d%d", &n, &m); SgT::n = n;for (int i = 1; i <= m; i++) {scanf("%d%d%d", l + i, r + i, q + i);add(l[i], r[i], q[i]);}for (int i = 1; i <= m; i++) {if (ask(l[i], r[i]) != q[i]) {puts("NO");exit(0);}}puts("YES");for (int i = 1; i < n; i++) printf("%d ", ask(i, i));printf("%d\n", ask(n, n));return 0;
}

转载于:https://www.cnblogs.com/Memory-of-winter/p/9753778.html

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