


codeforces Round #275(div2) DProblem Solution Report_html/css_WEB-ITnose
Jun 24, 2016 am 11:55 AM
D. Interesting Array
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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.
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
題目大意:
假設(shè)有n個非負(fù)數(shù),現(xiàn)在有m個限制,a[l] & a[l 1] & a[l 2] ... & a[r] = q。要求根據(jù)上述的限制,輸出符合要求的1~n個數(shù),如若不能則輸出“NO”。
解法:
? ? ? ?我們先挖掘題意,弄清楚題目給的已知條件和要我們輸出什么。
? ? ? ?a[l] & a[l 1] & a[l 2] ... & a[r] = q,這是每個限制的基本形式,由“&”我們可以得知,如若q中的某一個bit是1的話,則要求a[l]~a[r]中的那個bit位都為1。這個條件看似是限制,現(xiàn)在通過轉(zhuǎn)化,似乎可以成為我們的已知條件,即每一個a[i]中的必須要為1的bit。
? ? ? ?通過上述可知,我們得到每個a[i]的基本值,然后每一個限制是一個區(qū)間,很容易就想到了線段樹,對每一條限制進行查詢,看是否沖突,如若沖突則為"NO“,如若不沖突,則就按照a[i]的必須值來輸出即可。
代碼:
#include#include #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();}

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