[Codeforces 482B] Interesting Array

標籤：rest   codeforce   node   inline   ons   value   ORC   ==   col   

[題目連結]

         https://codeforces.com/contest/482/problem/B

[演算法]

        顯然 ， 當qi二進位表示下第j位為1時 ， [li,ri]中每個數二進位表示下的第j為也為1

        根據這個性質 ， 計算出要求的序列a， 然後用線段樹檢驗序列是否合法即可

        時間複雜度 ： O(NlogN)

[代碼]

         

#include<bits/stdc++.h>using namespace std;const int MAXN = 1e5 + 10;#define MAXLOG 31struct query{        int l , r , q;} a[MAXN];int n , m;int value[MAXN],cnt[MAXN];struct SegmentTree{        struct Node        {                int l , r , val;        } Tree[MAXN << 2];        inline void build(int index,int l,int r)        {                Tree[index].l = l;                Tree[index].r = r;                if (l == r)                 {                        Tree[index].val = value[l];                        return;                }                int mid = (l + r) >> 1;                build(index << 1,l,mid);                build(index << 1 | 1,mid + 1,r);                update(index);        }        inline void update(int index)        {                Tree[index].val = Tree[index << 1].val & Tree[index << 1 | 1].val;        }        inline int query(int index,int l,int r)        {                if (Tree[index].l == l && Tree[index].r == r) return Tree[index].val;                int mid = (Tree[index].l + Tree[index].r) >> 1;                if (mid >= r) return query(index << 1,l,r);                else if (mid + 1 <= l) return query(index << 1 | 1,l,r);                else return query(index << 1,l,mid) & query(index << 1 | 1,mid + 1,r);        }} T;template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }template <typename T> inline void read(T &x){    T f = 1; x = 0;    char c = getchar();    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘;    x *= f;}int main(){                read(n); read(m);        for (int i = 1; i <= m; i++)        {                read(a[i].l);                read(a[i].r);                read(a[i].q);        }        for (int i = 0; i < MAXLOG; i++)        {                for (int j = 0; j <= n; j++) cnt[j] = 0;                for (int j = 1; j <= m; j++)                {                        if (a[j].q & (1 << i))                        {                                cnt[a[j].l]++;                                cnt[a[j].r + 1]--;                                }                        }                 for (int j = 1; j <= n; j++)                 {                        cnt[j] += cnt[j - 1];                        if (cnt[j] > 0)                                value[j] |= (1 << i);                }        }         T.build(1,1,n);        for (int i = 1; i <= m; i++)        {                if (T.query(1,a[i].l,a[i].r) != a[i].q)                {                        printf("NO\n");                        return 0;                }        }        printf("YES\n");        for (int i = 1; i <= n; i++) printf("%d ",value[i]);        printf("\n");                return 0;    }

 

[Codeforces 482B] Interesting Array