Codeforces Round #275 (Div. 1) B. Interesting Array (线段树)

题目链接：http://codeforces.com/contest/482/problem/B

题意：给定一个长度为n个序列，要求满足m个条件：[L,R]区间内的异或和为x

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <queue>
#include <climits>
#include <set>
#include <stack>
#include <string>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
using namespace std;
typedef long long ll;
static const int MAX_N = 1e5 + 5;
static const int N = 1005;
static const ll Mod = 2009;
static const int inf = (1 << 30) - 1;
struct Node{
    int l, r, x;
}p[MAX_N];
int n;
int sum[MAX_N << 2], lazy[MAX_N << 2];
void push_up(int rt){
    sum[rt] = sum[rt << 1] & sum[rt << 1 | 1];
}
void push_down(int rt){
    if(lazy[rt]){
        sum[rt << 1] |= lazy[rt];
        sum[rt << 1 | 1] |= lazy[rt];
        lazy[rt << 1] |= lazy[rt];
        lazy[rt << 1 | 1] |= lazy[rt];
        lazy[rt] = 0;
    }
}
void update(int l, int r, int rt, int L, int R, int x){
    if(l >= L && r <= R){
        sum[rt] |= x;
        lazy[rt] |= x;
        return ;
    }
    push_down(rt);
    int m = l + r >> 1;
    if(m >= L) update(lson, L, R, x);
    if(R > m) update(rson, L, R, x);
    push_up(rt);
}
int query(int l, int r, int rt, int L, int R){
    if(l >= L && r <= R) return sum[rt];
    push_down(rt);
    int m = l + r >> 1;
    int ans = inf;
    if(m >= L) ans &= query(lson, L, R);
    if(R > m) ans &= query(rson, L, R);
    push_up(rt);
    return ans;
}
void output(int l, int r, int rt){
    if(l == r){
        printf("%d%c", sum[rt], l == n ? '\n' : ' ');
        return ;
    }
    push_down(rt);
    int m = l + r >> 1;
    output(lson);
    output(rson);
}
void solve(){
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    int m;
    scanf("%d%d", &n, &m);
    for(int i = 0; i < m; ++i){
        scanf("%d%d%d", &p[i].l, &p[i].r, &p[i].x);
        update(1, n, 1, p[i].l, p[i].r, p[i].x);
    }
    bool fg = true;
    for(int i = 0; i < m; ++i){
        if(p[i].x != query(1, n, 1, p[i].l, p[i].r)){
            fg = false;
            break;
        }
    }
    if(!fg) puts("NO");
    else{
        puts("YES");
        output(1, n, 1);
    }
}
int main() {
    solve();
    return 0;
}