https://codeforces.com/problemset/problem/482/B
题目描述
We'll call an array of nn non-negative integers a\[1\],a\[2\],...,a\[n\] interesting, if it meets mm constraints. The ii -th of the mm constraints consists of three integers l_{i}li , r_{i}ri , q_{i}qi ( 1<=l_{i}<=r_{i}<=n1<=li<=ri<=n ) meaning that value should be equal to q_{i}qi .
Your task is to find any interesting array of nn elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers xx and yy . In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
输入格式
We'll call an array of nn non-negative integers a\[1\],a\[2\],...,a\[n\] interesting, if it meets mm constraints. The ii -th of the mm constraints consists of three integers l_{i}li , r_{i}ri , q_{i}qi ( 1<=l_{i}<=r_{i}<=n1<=li<=ri<=n ) meaning that value should be equal to q_{i}qi .
Your task is to find any interesting array of nn elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers xx and yy . In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
输出格式
We'll call an array of nn non-negative integers a\[1\],a\[2\],...,a\[n\] interesting, if it meets mm constraints. The ii -th of the mm constraints consists of three integers l_{i}li , r_{i}ri , q_{i}qi ( 1<=l_{i}<=r_{i}<=n1<=li<=ri<=n ) meaning that value should be equal to q_{i}qi .
Your task is to find any interesting array of nn elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers xx and yy . In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
题意翻译
构建一个序列a,满足m条限制. 限制形如<l,r,q>: a[l]&a[l+1]&...&a[r-1]&a[r]=q;(此处'&'为位运算的and操作).
由 @Styx 提供翻译
输入输出样例
输入 #1复制
3 1 1 3 3
输出 #1复制
YES 3 3 3
输入 #2复制
3 2 1 3 3 1 3 2
输出 #2复制
NO
思路：运用线段树维护每个区间的&结果。
比如开始给一段区间a赋值p1,那么这个区间&的结果就是p1,而且区间内每个数字与p1二进制对应的1都要是1。
那现在再来一段区间b赋值p2,那么这个区间&的结果就是p2。那么a和b区间有交集的部分如何处理?有交集的部分就是p1|p2,这样分别在两个区间中的&值都能满足p1,p2.
那用线段树维护区间的|值，区间修改打lazy标记。
然后检查是否能够找到这样的序列呢？所有操作结束之后，对于每一个操作再进行一次查询，如果与操作不符，就是no，如果都是相符的就是yes，最后输出整个区间的所有元素就可以了。
不过有个奇怪的事为啥..query的res用LL就直接出问题...(雾
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
#define INF (1e18)-1
using namespace std;
const int maxn=1e5+100;
typedef long long LL;
LL l[maxn],r[maxn],p[maxn],n,m;
struct SeamentTree{
LL l,r,add,val;
}tree[maxn*4];
void pushup(LL p)
{
tree[p].val=tree[p*2].val&tree[p*2+1].val;
}
void spread(LL p)
{
if(tree[p].add)
{
tree[p*2].val|=tree[p].add;
tree[p*2+1].val|=tree[p].add;
tree[p*2].add|=tree[p].add;
tree[p*2+1].add|=tree[p].add;
tree[p].add=0;
}
}
void build(LL p,LL l,LL r)
{
tree[p].l=l;tree[p].r=r;
if(l==r){tree[p].val=0;return;}
LL mid=(l+r)/2;
build(p*2,l,mid);
build(p*2+1,mid+1,r);
pushup(p);
}
void change(LL p,LL l,LL r,LL d)
{
if(l<=tree[p].l&&r>=tree[p].r)
{
tree[p].val|=d;
tree[p].add|=d;
return;
}
spread(p);
LL mid=(tree[p].l+tree[p].r)/2;
if(l<=mid) change(p*2,l,r,d);
if(r>mid) change(p*2+1,l,r,d);
pushup(p);
}
LL query(LL p,LL l,LL r)
{
if(l<=tree[p].l&&r>=tree[p].r) return tree[p].val;
spread(p);
LL mid=(tree[p].l+tree[p].r)/2;
int res=INF;///为啥LL不过阿
// debug(res);
// LL ans=INF;
// debug(ans);
if(l<=mid) res &= query(p*2,l,r);
if(r>mid) res &= query(p*2+1,l,r);
return res;
}
bool judge()
{
for(LL i=1;i<=m;i++)
{
if(query(1,l[i],r[i])!=p[i]) return true;
}
cout<<"YES"<<endl;
for(LL i=1;i<n;i++){
cout<<query(1,i,i)<<" ";
}
cout<<query(1,n,n)<<endl;
return false;
}
int main(void)
{
cin.tie(0);std::ios::sync_with_stdio(false);
cin>>n>>m;
build(1,1,n);
for(LL i=1;i<=m;i++)
{
cin>>l[i]>>r[i]>>p[i];
change(1,l[i],r[i],p[i]);
}
if(judge()) cout<<"NO"<<endl;
return 0;
}