hdu 4638 Group 莫队算法
很裸的莫队， 就不多说了...
1 #include<bits/stdc++.h> 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; i<n; i++) 12 #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<m; 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<m; i++) { 58 x = q[i].l, y = q[i].r; 59 for (int i=x; i<L; 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<x; 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<m; i++) 91 printf("%d\n", ans[i]); 92 } 93 return 0; 94 }
