codeforces Round #275(div2) D解题报告_html/css_WEB-ITnose

D. Interesting Array

time limit per test

memory limit per test

input

output

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".

Input

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.

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Each of the next m lines contains three integers li, ri, qi (1?≤?li?≤?ri?≤?n, 0?≤?qi?230) describing the i-th limit.

Output

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.

Sample test(s)

input

3 11 3 3

output

YES3 3 3

input

3 21 3 31 3 2

output

NO

题目大意：

假设有n个非负数，现在有m个限制，a[l] & a[l+1] & a[l+2] ... & a[r] = q。要求根据上述的限制，输出符合要求的1~n个数，如若不能则输出“NO”。

解法：

       我们先挖掘题意，弄清楚题目给的已知条件和要我们输出什么。

       a[l] & a[l+1] & a[l+2] ... & a[r] = q，这是每个限制的基本形式，由“&”我们可以得知，如若q中的某一个bit是1的话，则要求a[l]~a[r]中的那个bit位都为1。这个条件看似是限制，现在通过转化，似乎可以成为我们的已知条件，即每一个a[i]中的必须要为1的bit。

       通过上述可知，我们得到每个a[i]的基本值，然后每一个限制是一个区间，很容易就想到了线段树，对每一条限制进行查询，看是否冲突，如若冲突则为"NO“，如若不冲突，则就按照a[i]的必须值来输出即可。

代码：

#include <cstdio>#include <cstring>#define Maxbit 29#define M_max 123456#define N_max 123456#define root 1, 1, nusing namespace std;const int noth = (1<<30)-1;int n, m;int l[M_max], r[M_max], q[M_max], a[N_max];int sum[N_max], tree[N_max*3];void build(int v, int l, int r) {	if (l == r) {		tree[v] = a[l];		return;	}	int ls = v<<1, rs = ls+1, mid = (l+r)>>1;	build(ls, l, mid);	build(rs, mid+1, r);	tree[v] = tree[ls] & tree[rs];}int query(int v, int l, int r, int ql, int qr) {	if (r < ql || l > qr)  return noth;	if (ql <= l && r <= qr)  return tree[v];	int ls = v<<1, rs = ls+1, mid = (l+r)>>1;	return query(ls, l, mid, ql, qr) & query(rs, mid+1, r, ql, qr);}void init() {	scanf("%d%d", &n, &m);	for (int i = 1; i <= m; i++)		scanf("%d%d%d", &l[i], &r[i], &q[i]);	for (int i = 0; i <= Maxbit; i++) {		memset(sum, 0, sizeof(sum));		for (int j = 1; j <= m; j++)			if ((q[j] >> i) & 1) {				sum[l[j]]++;				sum[r[j]+1]--;			}		for (int j = 1; j <= n; j++) {			sum[j] += sum[j-1];			if (sum[j] > 0)  a[j] |= 1 << i;		}	}	build(root);}void solve() {	for (int i = 1; i <= m; i++)		if (query(root, l[i], r[i]) != q[i]) {			printf("NO\n");			return;		}	printf("YES\n");	for (int i = 1; i <= n; i++)  printf("%d ", a[i]);	printf("\n");}int main() {	init();	solve();}