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：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<vector>
 6 
 7 using namespace std;
 8 
 9 const int maxn = 100010;
10 const int N = 550;
11 const int INF = 0x3f3f3f3f;
12 
13 int a[maxn], belong[maxn];
14 int L[N], R[N], add[N];
15 int block, num;
16 vector<int> vc[N];
17 
18 inline void build(int n){
19     block = sqrt(n);
20     num = ceil(n * 1.0 / block);
21     for(int i = 1; i <= n; i++){
22         belong[i] = (i - 1) / block + 1;
23         vc[belong[i]].push_back(a[i]);
24     }
25     for(int i = 1; i <= num; i++){
26         L[i] = R[i - 1] + 1;
27         R[i] = i * block;
28         sort(vc[i].begin(), vc[i].end());
29     }
30     R[num] = n;
31 }
32 
33 inline void update(int pos){
34     vc[pos].clear();
35     for(int i = L[pos]; i <= R[pos]; i++) vc[pos].push_back(a[i]);
36     sort(vc[pos].begin(), vc[pos].end());
37 }
38 
39 inline void modify(int l, int r, int c){
40     for(int i = l; i <= min(r, R[belong[l]]); i++) a[i] += c;
41     update(belong[l]);
42     if(belong[l] != belong[r]){
43         for(int i = L[belong[r]]; i <= r; i++) a[i] += c;
44         update(belong[r]);
45     }
46     for(int i = belong[l] + 1; i < belong[r]; i++) add[i] += c;
47 }
48 
49 inline int query(int l, int r, int c){
50     int ans = -INF;
51     for(int i = l; i <= min(r, R[belong[l]]); i++)
52         if(a[i] + add[belong[i]] < c) ans = max(ans, a[i] + add[belong[i]]);
53     if(belong[l] != belong[r]){
54         for(int i = L[belong[r]]; i <= r; i++)
55             if(a[i] + add[belong[i]] < c) ans = max(ans, a[i] + add[belong[i]]);
56     }
57     for(int i = belong[l] + 1; i < belong[r]; i++){
58         if(vc[i][0] + add[i] >= c) continue;
59         int k = lower_bound(vc[i].begin(), vc[i].end(), c - add[i]) - vc[i].begin();
60         ans = max(ans, vc[i][k - 1] + add[i]);
61     }
62     return ans;
63 }
64 
65 int main(){
66     int n;
67     scanf("%d", &n);
68     for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
69     build(n);
70     for(int i = 1; i <= n; i++){
71         int opt, l, r, c;
72         scanf("%d%d%d%d", &opt, &l, &r, &c);
73         if(opt == 0) modify(l, r, c);
74         else{
75             int k = query(l, r, c);
76             if(k == -INF) printf("-1\n");
77             else printf("%d\n", k);
78         }
79     }
80     return 0;
81 }

AC代码