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Codeforces E. Interesting Array(线段树)

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题目描述:

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 them 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 integersa[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.
Examples
input

3 1
1 3 3

output

YES
3 3 3

input

3 2
1 3 3
1 3 2

output

NO

思路:

跟区间相关的问题可以考虑用线段树。这道题是要通过一次次的限制,把数组中的数构造出来。比如第l个到第r个数的&结果为q,即在二进制表示上,如果某一位并的结果为一,那么该区间上所有数的该位全都为一,若某一位结果为零,那么该区间上至少有一个数的该位为0。

刚开始:全为一时知道区间上所有数在这一位上为一,并为零时不知道区间上那一个数的这一位为零。想构造二维向量,将数组中的数用二进制表示,暴力解出可能的结果。但好像操作起来挺复杂(编程实现有点困难,对我这个蒟蒻来说),也可能会超时。

最后:考虑线段树,来结合问题看一看。在区间【l,r】上,假设并的结果为3(样例1),那么区间上的每一个数(用sum数组来实现线段树)为线段树的叶节点,刚开始sum[i]=0,在一次次的限制条件下,用sum[k]|=q[i],(q[i]表示并的结果),即用或操作我们可以构造出sum[k]来满足一条条的限制条件,当所有的限制条件过完了后,在来查询一个个限制条件所给的区间上数的并是否依然等于q[i],因为在刚刚构造过程中前面满足的条件可能会因为后面的限制条件而改变,改变就说明出现了两个矛盾的限制条件,因为对不矛盾的限制条件来说,他们构造的数应是要么相等,要么构造的数不一是同一个。

在求区间和的线段树模板上稍作改动。具体的,把

pushup(k){sum[k] = sum[k<<1]+sum[k<<1|1]的加改为并,在询问操作中的返回结果思考了有点久,这跟和好像不太一样,注意题目中的数据范围(小于2^30),就用I=2^30-1,代替ans += ···为I &=···来返回结果。

小心的是pushdown的时刻,返回时怎么返回结果,还有一点:2<<30-1你以为是2^30-1?错了!注意运算符的优先级,写成(2<<30)-1

知识点:线段树

代码:

 1 #include <iostream>
 2 #define max_n 100005
 3 using namespace std;
 4 int sum[max_n<<2];
 5 int lz[max_n<<2];
 6 int l[max_n];
 7 int r[max_n];
 8 int q[max_n];
 9 int n;
10 int m;
11 int I;
12 void pushup(int k)
13 {
14     sum[k] = sum[k<<1]&sum[k<<1|1];
15 }
16 void pushdown(int k,int ln,int rn)
17 {
18     if(lz[k])
19     {
20         sum[k<<1] |= lz[k];
21         sum[k<<1|1] |= lz[k];
22         lz[k<<1] |= lz[k];
23         lz[k<<1|1] |= lz[k];
24         lz[k] = 0;
25     }
26 }
27 void build(int k,int l,int r)
28 {
29     lz[k] = 0;
30     if(l==r)
31     {
32         sum[k]=0;
33         return;
34     }
35     int mid = (l+r)>>1;
36     build(k<<1,l,mid);
37     build(k<<1|1,mid+1,r);
38     pushup(k);
39 }
40 void update(int k,int L,int R,int l,int r,int val)
41 {
42     if(L<=l&&r<=R)
43     {
44         sum[k] |= val;
45         lz[k] |= val;
46         return;
47     }
48     int mid = (l+r)/2;
49     int ln = mid-l+1;
50     int rn = r-mid;
51     pushdown(k,ln,rn);
52     if(L<=mid) update(k<<1,L,R,l,mid,val);
53     if(mid<R) update(k<<1|1,L,R,mid+1,r,val);
54     pushup(k);
55 }
56 int query(int k,int L,int R,int l,int r)
57 {
58     if(L<=l&&r<=R)
59     {
60         //cout << "k " << k << " " << sum[k] << endl;
61         return sum[k];
62     }
63     int mid = (l+r)>>1;
64     int ln = mid-l+1;
65     int rn = r-mid;
66     pushdown(k,ln,rn);
67     long long I = (long long)(2<<30) - 1;
68     if(L<=mid) I = I&query(k<<1,L,R,l,mid);
69     //cout << "qian " << I << endl;
70     if(mid<R) I = I&query(k<<1|1,L,R,mid+1,r);
71     pushup(k);
72     //cout << "hou " << I << endl;
73     return I;
74 }
75 int main()
76 {
77     cin >> n >> m;
78     build(1,1,n);
79     for(int i = 0;i<m;i++)
80     {
81         cin >> l[i] >> r[i] >> q[i];
82         update(1,l[i],r[i],1,n,q[i]);
83     }
84     for(int i = 0;i<m;i++)
85     {
86         int ans = query(1,l[i],r[i],1,n);
87         if(ans!=q[i])
88         {
89             cout << "NO" << endl;
90             return 0;
91         }
92     }
93     cout << "YES" << endl;
94     for(int i = 1;i<=n;i++)
95     {
96         cout << query(1,i,i,1,n) << " ";
97     }
98     return 0;
99 }

 

标签:interesting,int,Interesting,线段,Codeforces,sum,区间,Array,array
来源: https://www.cnblogs.com/zhanhonhao/p/11221221.html