分块

I Hate It

HDU - 1754
 
 
 1 //fen kuai
 2 #include <bits/stdc++.h>
 3 using namespace std;
 4 const int maxn = 200010;
 5 const int inf = 0x3f3f3f3f;
 6 int n, m, SZ;
 7 int a[maxn], bc[maxn], Max[maxn];
 8 
 9 void update(int p, int v){
10     int tp = (p - 1) / SZ + 1;
11     a[p] = v;
12     int temp = -inf;
13     for(int i = (tp - 1) * SZ + 1; i <= tp * SZ; i++){
14         temp = max(temp, a[i]);
15     }
16     Max[tp] = temp;
17 }
18 
19 int query(int l, int r){
20     int ans = -inf;
21     int pl = (l - 1) / SZ + 1;
22     for(int i = l; i <= min(r, pl * SZ); i++){
23         ans = max(ans, a[i]);
24     }
25     int pr = (r - 1) / SZ + 1;
26     if(pl != pr){
27         for(int i = (pr - 1) * SZ + 1; i <= r; i++){
28             ans = max(ans, a[i]);
29         }
30     }
31     for(int i = pl + 1; i < pr; i++){
32         ans = max(ans, Max[i]);
33     }
34     return ans;
35 }
36 
37 int main(){
38     while(scanf("%d %d", &n, &m) != EOF){
39         memset(Max, -inf, sizeof Max);
40         SZ = sqrt(n * 1.0);
41         for(int i = 1; i <= n; i++){
42             scanf("%d", &a[i]);
43             bc[i] = (i - 1) / SZ + 1;
44             Max[bc[i]] = max(Max[bc[i]], a[i]);
45         }
46         while(m--){
47             char op[4];
48             int l, r;
49             scanf("%s %d %d", op, &l, &r);
50             if(op[0] == 'Q') {
51                 printf("%d\n", query(l, r));
52             }else{
53                 update(l, r);
54             }
55         }
56     }
57     return 0;
58 }
View Code

 

 

A Simple Problem with Integers

POJ - 3468

 

 1 //fen kuai
 2 #include <iostream>
 3 #include <cmath>
 4 #include <cstdio>
 5 #include <cstring>
 6 using namespace std;
 7 #define LL long long
 8 const int maxn = 100010;
 9 LL a[maxn], atag[maxn];
10 LL sum[maxn];
11 int n, m, SZ;
12 
13 LL query(int l, int r){
14     LL ans = 0;
15     int pl = (l - 1) / SZ + 1;
16     int pr = (r - 1) / SZ + 1;
17     for(int i = l; i <= min(r, pl * SZ); i++){
18         ans += a[i] + atag[pl];
19     }
20     if(pl != pr){
21         for(int i = (pr - 1) * SZ + 1; i <= r; i++){
22             ans += a[i] + atag[pr];
23         }
24     }
25     for(int i = pl + 1; i < pr; i++){
26         ans += sum[i];
27     }
28     return ans;
29 }
30 
31 void update(int l, int r, LL c){
32     int pl = (l - 1) / SZ + 1;
33     int pr = (r - 1) / SZ + 1;
34     for(int i = l; i <= min(r, pl * SZ); i++){
35         a[i] += c;
36         sum[pl] += c;
37     }
38     if(pl != pr){
39         for(int i = (pr - 1) * SZ + 1; i <= r; i++){
40             a[i] += c;
41             sum[pr] += c;
42         }
43     }
44     for(int i = pl + 1; i < pr; i++){
45         sum[i] += c * SZ;
46         atag[i] += c;
47     }
48 }
49 
50 int main(){
51     while(scanf("%d %d", &n, &m) != EOF){
52         SZ = sqrt(n * 1.0);
53         int j = 1;
54         for(int i = 1; i <= n; i++){
55             sum[i] = 0;
56             atag[i] = 0;
57             scanf("%lld", &a[i]);
58             sum[j] += a[i];
59             if(i % SZ == 0) j++;
60         }
61         while(m--){
62             char op[4];
63             int x, y;
64             LL c;
65             scanf("%s %d %d", op, &x, &y);
66             if(op[0] == 'Q'){
67                 printf("%lld\n", query(x, y));
68             }else{
69                 scanf("%lld", &c);
70                 update(x, y, c);
71             }
72         }
73     }    
74     return 0;
75 }
View Code

 

 

Paint The Wall

HDU - 4391

参考：https://www.cnblogs.com/swm8023/archive/2012/09/23/2698780.html

带懒标记的分块

 1 // lazy 分块
 2 #include <bits/stdc++.h>
 3 using namespace std;
 4 const int maxn = 100010;
 5 
 6 int c[maxn], ov[maxn];
 7 map<int, int> mp[maxn];
 8 int n, m, SZ;
 9 
10 void pushdown(int id){
11     if(ov[id] != -1){
12         for(int i = id * SZ; i < min((id + 1) * SZ, n); i++) c[i] = ov[id];
13         mp[id].clear();
14         mp[id][ov[id]] = min((id + 1) * SZ, n) - id * SZ;
15         ov[id] = -1;
16     }
17 }
18 
19 void update(int l, int r, int z){
20     int pl = l / SZ, pr = r / SZ;
21     pushdown(pl);
22     for(int i = l; i <= min(r, (pl + 1) * SZ - 1); i++){
23         mp[pl][c[i]]--;
24         c[i] = z;
25         mp[pl][z]++;
26     }
27     if(pl != pr){
28         pushdown(pr);
29         for(int i = pr * SZ; i <= r; i++){
30             mp[pr][c[i]]--;
31             c[i] = z;
32             mp[pr][z]++;
33         } 
34     }
35     for(int i = pl + 1; i < pr; i++) ov[i] = z;
36 }
37 int query(int l, int r, int z){
38     int res = 0;
39     int pl = l / SZ, pr = r / SZ;
40     pushdown(pl);
41     for(int i = l; i <= min(r, (pl + 1) * SZ - 1); i++) if(c[i] == z) res++;
42     if(pl != pr) {
43         pushdown(pr);
44         for(int i = pr * SZ; i <= r; i++) if(c[i] == z) res++;
45     }
46     for(int i = pl + 1; i < pr; i++){
47         if(ov[i] == -1){
48             if(mp[i].find(z) != mp[i].end()) res += mp[i][z];
49         }else if(ov[i] == z) res += min((i + 1) * SZ, n) - i * SZ;
50     } 
51     return res;
52 }
53 int main(){
54     while(scanf("%d %d", &n, &m) != EOF){
55         SZ = sqrt(n * 1.0);
56         memset(ov, -1, sizeof ov);
57         for(int i = 0; i < n; i++) mp[i].clear();
58         for(int i = 0; i < n; i++){
59             scanf("%d", &c[i]);
60             mp[i / SZ][c[i]]++; 
61         }
62         while(m--){
63             int op, l, r, z;
64             scanf("%d %d %d %d", &op, &l, &r, &z);
65             if(op == 1){
66                 update(l, r, z);
67             }else{
68                 printf("%d\n", query(l, r, z));
69             }
70         }
71     }
72     return 0;
73 }
View Code

 

posted @ 2018-01-24 19:09  yijiull  阅读(139)  评论(0)  编辑  收藏  举报