【线段树与区间或】区间或

最新推荐文章于 2024-08-11 00:01:12 发布

ljq真的不能再吃了

最新推荐文章于 2024-08-11 00:01:12 发布

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分类专栏： ACM DS-线段树

本文链接：https://blog.youkuaiyun.com/heucodesong/article/details/89324270

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DS-线段树

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这篇博客探讨如何使用线段树解决区间异或问题。根据题目，给定一个整数序列和Q个区间查询，需要判断能否构建线段树使得每个查询区间内的异或值等于特定目标值p[i]。采用贪心策略，对于1进行按位与所有1的操作，对于0则按位与0。最后比较计算结果与p[i]是否一致。

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给你一个整数序列让你满足Q个关系 l - r 区间内异或的数值为 p[i]
能则构造一个线段树不能则输出NO
我们用贪心的思路去是1就与上所有1 是0 就与上0
最后看异或出来的值是否和p[i]相等即可

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = (1<<30)-1;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 100025;
int s[MAX_N<<2],col[MAX_N<<2],l[MAX_N],r[MAX_N],p[MAX_N];
void down(int rt,int l,int r)
{
    if(col[rt]!=inf)
    {
        col[rt<<1] &= col[rt];
        col[rt<<1|1] &= col[rt];
        s[rt<<1] &= s[rt];
        s[rt<<1|1] &= s[rt];
        col[rt] = inf;
    }
}
void build(int rt,int l,int r)
{
    s[rt] = col[rt] = inf;
    if(l==r)
    {
        return ;
    }
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
}
void update(int rt,int l,int r,int x,int y,int v)
{
    if(x<=l&&r<=y)
    {
        s[rt]&=v;
        col[rt]&=v;
        return ;
    }
    down(rt,l,r);
    if(x<=mid) update(rt<<1,l,mid,x,y,v);
    if(mid<y) update(rt<<1|1,mid+1,r,x,y,v);
}
int query(int rt,int l,int r,int x,int y)
{
    if(x<=l&&r<=y)
        return s[rt];
    down(rt,l,r);
    int res = 0;
    if(x<=mid) res |= query(rt<<1,l,mid,x,y);
    if(mid<y) res |= query(rt<<1|1,mid+1,r,x,y);
    return res;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,Q;
    scanf("%d%d",&n,&Q);
    build(1,1,n);
    for(int i = 1;i<=Q;++i)
    {
        scanf("%d%d%d",&l[i],&r[i],&p[i]);
        update(1,1,n,l[i],r[i],p[i]);
    }
    for(int i = 1;i<=Q;++i)
    {
        if(query(1,1,n,l[i],r[i])!=p[i])
        {
            printf("No\n");
            return 0;
        }
    }
    printf("Yes\n");
    for(int i = 1;i<=n;++i)
    {
        i==n?printf("%d\n",query(1,1,n,i,i)):printf("%d ",query(1,1,n,i,i));
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}