hdu 4638 Group 莫队算法

题目链接

很裸的莫队， 就不多说了...

 1 #include
 2 using namespace std;
 3 #define pb(x) push_back(x)
 4 #define ll long long
 5 #define mk(x, y) make_pair(x, y)
 6 #define lson l, m, rt<<1
 7 #define mem(a) memset(a, 0, sizeof(a))
 8 #define rson m+1, r, rt<<1|1
 9 #define mem1(a) memset(a, -1, sizeof(a))
10 #define mem2(a) memset(a, 0x3f, sizeof(a))
11 #define rep(i, a, n) for(int i = a; i12 #define ull unsigned long long
13 typedef pair<int, int> pll;
14 const double PI = acos(-1.0);
15 const double eps = 1e-8;
16 const int mod = 1e9+7;
17 const int inf = 1061109567;
18 const int maxn = 1e5+5;
19 int vis[maxn];
20 int a[maxn], ans[maxn];
21 
22 struct node {
23     int l, r, block, id;
24     bool operator < (node x) const
25     {
26         if(block == x.block)
27             return r < x.r;
28         return block < x.block;
29     }
30 } q[maxn];
31 int main() {
32     int t, n, m;
33     cin>>t;
34     while (t--) {
35         scanf("%d%d", &n, &m);
36         int BLOCK = sqrt(n*1.0);
37         for(int i=1; i<=n; i++)
38             scanf("%d", &a[i]);
39         for(int i=0; i) {
40             scanf("%d%d", &q[i].l, &q[i].r);
41             q[i].block = q[i].l/BLOCK;
42             q[i].id = i;
43         }
44         sort(q, q+m);
45         mem(vis);
46         int tmp = 0;
47         int x = q[0].l, y = q[0].r;
48         for(int i=x; i<=y; i++) {
49             vis[a[i]] = 1;
50             if (vis[a[i]-1] && vis[a[i]+1])
51                 tmp--;
52             else if (!vis[a[i]-1] && !vis[a[i]+1])
53                 tmp++;
54         }
55         ans[q[0].id] = tmp;
56         int L = x, R = y;
57         for(int i=1; i) {
58             x = q[i].l, y = q[i].r;
59             for (int i=x; i) {
60                 vis[a[i]] = 1;
61                 if (vis[a[i]-1] && vis[a[i]+1])
62                     tmp--;
63                 else if (!vis[a[i]-1] && !vis[a[i]+1])
64                     tmp++;
65             }
66             for (int i=R+1; i<=y; i++) {
67                 vis[a[i]] = 1;
68                 if (vis[a[i]-1] && vis[a[i]+1])
69                     tmp--;
70                 else if (!vis[a[i]-1] && !vis[a[i]+1])
71                     tmp++;
72             }
73             for (int i=L; i) {
74                 vis[a[i]] = 0;
75                 if (vis[a[i]-1] && vis[a[i]+1])
76                     tmp++;
77                 else if (!vis[a[i]-1] && !vis[a[i]+1])
78                     tmp--;
79             }
80             for (int i=y+1; i<=R; i++) {
81                 vis[a[i]] = 0;
82                 if (vis[a[i]-1] && vis[a[i]+1])
83                     tmp++;
84                 else if (!vis[a[i]-1] && !vis[a[i]+1])
85                     tmp--;
86             }
87             ans[q[i].id] = tmp;
88             L = x, R = y;
89         }
90         for(int i=0; i)
91             printf("%d\n", ans[i]);
92     }
93     return 0;
94 }

 

转载于:https://www.cnblogs.com/yohaha/p/5071692.html