Codeforces 482b interesting Array Construction + line segment tree determination is feasible

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Question:

Construct an N-long sequence with M restrictions:

Each limit [L, R] q

The sequence must satisfy all numbers in the range [L, R]. The result is Q.

Ideas:

Construct it directly, and then determine whether it is feasible ..

#include <stdio.h>#include <cstring>#include <iostream>#include <map>template <class T>inline bool rd(T &ret) {    char c; int sgn;    if(c=getchar(),c==EOF) return 0;    while(c!='-'&&(c<'0'||c>'9')) c=getchar();    sgn=(c=='-')?-1:1;    ret=(c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');    ret*=sgn;    return 1;}template <class T>inline void pt(T x) {    if (x <0) {        putchar('-');        x = -x;    }    if(x>9) pt(x/10);    putchar(x%10+'0');}using namespace std;typedef unsigned long long ull;#define L(x) tree[x].l#define R(x) tree[x].r#define Lson(x) (x<<1)#define Rson(x) (x<<1|1)#define V(x) tree[x].valconst int N = 100005;int hehe;int haha;struct node{    int l, r, val;}tree[N<<2];int a[N];void push_up(int id){    V(id) = V(Lson(id)) & V(Rson(id));}void build(int l, int r, int id){    L(id) = l ; R(id) = r;    if(l == r){        V(id) = a[l];        return ;    }    int mid = (l+r)>>1;    build(l, mid, Lson(id)); build(mid+1, r, Rson(id));    push_up(id);}int query(int l, int r, int id){    if(l == L(id) && R(id) == r){        return V(id);    }    int mid = (L(id) + R(id)) >> 1;    if(mid < l)        return query(l, r, Rson(id));    else if(r <= mid)        return query(l, r, Lson(id));    else return query(l, mid, Lson(id)) & query(mid+1, r, Rson(id));}int n, m;struct Q{    int l, r, q;}q[N];int o[N][32], t[N][32];void put(){    for(int i = 1; i <= n; i++)    {        pt(a[i]);        i == n ? puts("") : putchar(' ');    }}bool work(){    int now = 0;    int b[35] = {0};    for(int i = 1; i <= n; i++)    {        for(int j = 0; j < 31; j++)            if(o[i][j])            {                now |= (1<<j);                b[j]+=o[i][j];            }        a[i] = now        for(int j = 0; j < 31; j++)            if(t[i][j])            {                b[j]+=t[i][j];                if(b[j]==0)                now &= ~(1<<j);            }    }    build(1, n, 1);    for(int i = 1; i <= m; i++) {        if(query(q[i].l, q[i].r, 1) != q[i].q)return false;    }    return true;}bool input(){    memset(o, 0, sizeof o);    memset(t, 0, sizeof t);    for(int i = 1; i <= m; i++){        scanf("%d %d %d", &q[i].l, &q[i].r, &q[i].q);        for(int j = 0; j < 31; j++)            if(q[i].q&(1<<j))            {                o[q[i].l][j] ++;                t[q[i].r][j] --;            }    }    return true;}int main(){    while(cin>>n>>m){        input();        if(work()){            puts("YES");            put();        }        else puts("NO");    }    return 0;}

Codeforces 482b interesting Array Construction + line segment tree determination is feasible