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.

Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
一個(gè)構(gòu)造題。給m個(gè)操作，使得l到r的值與起來(lái)是q。要求構(gòu)造出這樣的一個(gè)數(shù)組。
我們要求一個(gè)區(qū)間內(nèi)的數(shù)與起來(lái)都是q，那么這個(gè)區(qū)間內(nèi)的數(shù)的二進(jìn)制位上都得是1。那么我們|（或）起來(lái)，就相當(dāng)于把所有的位上都變成1。最后再與起來(lái)。如果是0，就不用管了。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=1e5+100; struct node{int l;int r;int v;int lazy; }p[maxx<<2]; int lr[maxx],rr[maxx],vr[maxx]; int n,m;inline void pushup(int cur) {p[cur].v=p[cur<<1].v&p[cur<<1|1].v; } inline void pushdown(int cur) {if(p[cur].lazy){p[cur<<1].lazy|=p[cur].lazy;p[cur<<1|1].lazy|=p[cur].lazy;p[cur<<1].v|=p[cur].lazy;p[cur<<1|1].v|=p[cur].lazy;p[cur].lazy=0;} } inline void build(int l,int r,int cur) {p[cur].l=l;p[cur].r=r;p[cur].lazy=p[cur].v=0;if(l==r) return ;int mid=l+r>>1;build(l,mid,cur<<1);build(mid+1,r,cur<<1|1);pushup(cur); } inline void update(int l,int r,int x,int cur) {int L=p[cur].l;int R=p[cur].r;if(l<=L&&R<=r){p[cur].lazy|=x;p[cur].v|=x;return ;}pushdown(cur);int mid=L+R>>1;if(r<=mid) update(l,r,x,cur<<1);else if(l>mid) update(l,r,x,cur<<1|1);else {update(l,mid,x,cur<<1);update(mid+1,r,x,cur<<1|1);}pushup(cur); } inline int query(int l,int r,int cur) {int L=p[cur].l;int R=p[cur].r;if(l<=L&&R<=r) return p[cur].v;pushdown(cur);int mid=L+R>>1;if(r<=mid) return query(l,r,cur<<1);else if(l>mid) return query(l,r,cur<<1|1);else return query(l,mid,cur<<1)&query(mid+1,r,cur<<1|1); } int main() {while(scanf("%d%d",&n,&m)!=EOF){build(1,n,1);for(int i=1;i<=m;i++){scanf("%d%d%d",&lr[i],&rr[i],&vr[i]);update(lr[i],rr[i],vr[i],1);}int flag=1;for(int i=1;i<=m;i++){if(query(lr[i],rr[i],1)!=vr[i]){flag=0;break;}}if(flag) {puts("YES");for(int i=1;i<=n;i++)printf("%d ",query(i,i,1));printf("\n");}else puts("NO");}return 0; }

努力加油a啊，(^o)/~

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