LOJ 6279 数列分块入门3

嗯...

题目链接：https://loj.ac/problem/6279

这道题在分块的基础上用vc数组记录，然后最后分三块，两边暴力枚举找前驱，中间lower_bound找前驱。

AC代码：

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<vector>

 using namespace std;

 const int maxn = ;
 const int N = ;
 const int INF = 0x3f3f3f3f;

 int a[maxn], belong[maxn];
 int L[N], R[N], add[N];
 int block, num;
 vector<int> vc[N];

 inline void build(int n){
     block = sqrt(n);
     num = ceil(n * 1.0 / block);
     for(int i = ; i <= n; i++){
         belong[i] = (i - ) / block + ;
         vc[belong[i]].push_back(a[i]);
     }
     for(int i = ; i <= num; i++){
         L[i] = R[i - ] + ;
         R[i] = i * block;
         sort(vc[i].begin(), vc[i].end());
     }
     R[num] = n;
 }

 inline void update(int pos){
     vc[pos].clear();
     for(int i = L[pos]; i <= R[pos]; i++) vc[pos].push_back(a[i]);
     sort(vc[pos].begin(), vc[pos].end());
 }

 inline void modify(int l, int r, int c){
     for(int i = l; i <= min(r, R[belong[l]]); i++) a[i] += c;
     update(belong[l]);
     if(belong[l] != belong[r]){
         for(int i = L[belong[r]]; i <= r; i++) a[i] += c;
         update(belong[r]);
     }
     for(int i = belong[l] + ; i < belong[r]; i++) add[i] += c;
 }

 inline int query(int l, int r, int c){
     int ans = -INF;
     for(int i = l; i <= min(r, R[belong[l]]); i++)
         if(a[i] + add[belong[i]] < c) ans = max(ans, a[i] + add[belong[i]]);
     if(belong[l] != belong[r]){
         for(int i = L[belong[r]]; i <= r; i++)
             if(a[i] + add[belong[i]] < c) ans = max(ans, a[i] + add[belong[i]]);
     }
     for(int i = belong[l] + ; i < belong[r]; i++){
         if(vc[i][] + add[i] >= c) continue;
         int k = lower_bound(vc[i].begin(), vc[i].end(), c - add[i]) - vc[i].begin();
         ans = max(ans, vc[i][k - ] + add[i]);
     }
     return ans;
 }

 int main(){
     int n;
     scanf("%d", &n);
     for(int i = ; i <= n; i++) scanf("%d", &a[i]);
     build(n);
     for(int i = ; i <= n; i++){
         int opt, l, r, c;
         scanf("%d%d%d%d", &opt, &l, &r, &c);
         if(opt == ) modify(l, r, c);
         else{
             int k = query(l, r, c);
             if(k == -INF) printf("-1\n");
             else printf("%d\n", k);
         }
     }
     return ;
 }

AC代码