标签：tle com update stat graph cond tin integer this
We‘ll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value 
should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn‘t exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n](0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn‘t exist, print "NO" (without the quotes) in the single line.
3 1
1 3 3
YES
3 3 3
3 2
1 3 3
1 3 2
NO
算法：线段树
题解：直接用线段树版子，更新用按位或来修改，用按位与来向上更新，要快点可以打个lazy数组。
#include <iostream> #include <cstdio> #include <algorithm> #include <set> #include <map> #include <vector> using namespace std; #define INF 0x3f3f3f3f typedef long long ll; const int maxn = 1e5+7; struct node { int l, r, s; }tree[maxn << 2]; struct queue { int l, r, w; }Q[maxn]; int n, m; int lazy[maxn << 2]; void build(int root, int l, int r) { tree[root].l = l; tree[root].r = r; tree[root].s = 0; if(l == r) { return; } int mid = (l + r) >> 1; build(root << 1, l, mid); build(root << 1 | 1, mid + 1, r); } void pushup(int root) { tree[root].s = tree[root << 1].s & tree[root << 1 | 1].s; } void pushdown(int root) { if(lazy[root]) { tree[root << 1].s |= lazy[root]; tree[root << 1 | 1].s |= lazy[root]; lazy[root << 1] |= lazy[root]; lazy[root << 1 | 1] |= lazy[root]; lazy[root] = 0; } } void update(int root, int x, int y, int val) { int l = tree[root].l; int r = tree[root].r; if(x <= l && r <= y) { tree[root].s |= val; lazy[root] |= val; return; } pushdown(root); int mid = (l + r) >> 1; if(x <= mid) { update(root << 1, x, y, val); } if(y > mid) { update(root << 1 | 1, x, y, val); } pushup(root); } int queue(int root, int x, int y) { int l = tree[root].l; int r = tree[root].r; if(x <= l && r <= y) { return tree[root].s; } pushdown(root); int mid = (l + r) >> 1; int ans = (1 << 30) - 1; if(x <= mid) { ans &= queue(root << 1, x, y); } if(y > mid) { ans &= queue(root << 1 | 1, x, y); } pushdown(root); return ans; } int main() { scanf("%d %d", &n, &m); build(1, 1, n); for(int i = 1; i <= m; i++) { scanf("%d %d %d", &Q[i].l, &Q[i].r, &Q[i].w); update(1, Q[i].l, Q[i].r, Q[i].w); } for(int i = 1; i <= m; i++) { if(queue(1, Q[i].l, Q[i].r) != Q[i].w) { //如果其中有一个不满足条件的话，就输出NO printf("NO\n"); return 0; } } printf("YES\n"); for(int i = 1; i <= n; i++) { printf("%d%c", queue(1, i, i), " \n"[i == n]); } return 0; }
标签：tle com update stat graph cond tin integer this
原文地址：https://www.cnblogs.com/buhuiflydepig/p/11345810.html
