BZOJ1012，特点是只往后加所以可用单调栈。亦可无脑线段树。
1 #include <bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 5 int m; 6 ll D, last; 7 ll q[200005]; 8 int id[200005], tail, t; 9 10 inline void add(ll x) { 11 while (tail && q[tail] <= x) 12 tail--; 13 q[++tail] = x; 14 id[tail] = ++t; 15 } 16 17 inline ll query(ll x) { 18 ll k = t - x + 1; 19 int pos = lower_bound(id+1, id+1+tail, k) - id; 20 return q[pos]; 21 } 22 23 int main() { 24 cin >> m >> D; 25 while (m--) { 26 char ch[3]; 27 ll a; 28 scanf("%s%lld", ch, &a); 29 if (ch[0] == 'A') { 30 add((a + last) % D); 31 } else { 32 printf("%lld ", last = query(a)); 33 } 34 } 35 return 0; 36 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int maxm = 200005; 5 int M, D, last, n; 6 struct Seg { 7 int l, r, maxx; 8 }t[maxm << 2]; 9 10 void build(int l, int r, int p) { 11 t[p].l = l, t[p].r = r; 12 if (l == r) { 13 t[p].maxx = 0; 14 return; 15 } 16 int mid = (l + r) >> 1; 17 int ls = p << 1, rs = p << 1 | 1; 18 build(l, mid, ls); 19 build(mid+1, r, rs); 20 t[p].maxx = max(t[ls].maxx, t[rs].maxx); 21 } 22 23 void Add(int l, int k, int p) { 24 if (t[p].l == t[p].r) { 25 t[p].maxx = k; 26 return; 27 } 28 int mid = (t[p].l + t[p].r) >> 1; 29 int ls = p << 1, rs = p << 1 | 1; 30 if (l <= mid) Add(l, k, ls); 31 else Add(l, k, rs); 32 t[p].maxx = max(t[ls].maxx, t[rs].maxx); 33 } 34 35 int Query(int l, int r, int p) { 36 if (l <= t[p].l && t[p].r <= r) { 37 return t[p].maxx; 38 } 39 int mid = (t[p].l + t[p].r) >> 1; 40 int ls = p << 1, rs = p << 1 | 1; 41 int ret = 0; 42 if (l <= mid) ret = max(ret, Query(l, r, ls)); 43 if (mid < r) ret = max(ret, Query(l, r, rs)); 44 return ret; 45 } 46 47 int main() { 48 scanf("%d%d", &M, &D); 49 build(1, M, 1); 50 while (M--) { 51 char ch[3]; 52 int x; 53 scanf("%s%d", ch, &x); 54 if (ch[0] == 'A') { 55 ++n; 56 Add(n, (x+last)%D, 1); 57 } else { 58 printf("%d ", last = Query(n-x+1, n, 1)); 59 } 60 } 61 }
BZOJ2724，分块框架下的数据预处理技巧。
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <cctype> 5 #include <iostream> 6 #include <algorithm> 7 #include <vector> 8 #include <map> 9 #define ri readint() 10 #define gc getchar() 11 #define rep(i, l, r) for (int i = l; i <= r; i++) 12 using namespace std; 13 14 inline int readint() { 15 int x = 0, s = 1, c = gc; 16 while (c <= 32) c = gc; 17 if (c == '-') s = -1, c = gc; 18 for (; isdigit(c); c = gc) 19 x = x * 10 + c - 48; 20 return x * s; 21 } 22 23 const int maxn = 4e4 + 5; 24 int n, m, T, last; 25 int a[maxn], val[maxn], id; 26 map<int, int> mp; 27 vector<int> v[maxn]; 28 int block[maxn], f[205][205]; 29 30 void pre(int x) { 31 int cnt[maxn]; 32 memset(cnt, 0, sizeof(cnt)); 33 int maxx = 0, k = 0; 34 rep(i, (x - 1) * T + 1, n) { 35 int t = a[i]; 36 cnt[t]++; 37 if (cnt[t] > maxx || (cnt[t] == maxx && val[t] < val[k])) { 38 maxx = cnt[k = t]; 39 } 40 f[x][block[i]] = k; 41 } 42 } 43 44 int Query(int ql, int qr, int k) { 45 return upper_bound(v[k].begin(), v[k].end(), qr) - lower_bound(v[k].begin(), v[k].end(), ql); 46 } 47 48 int Query(int ql, int qr) { 49 int res = f[block[ql]+1][block[qr]-1]; 50 int maxx = Query(ql, qr, res); 51 rep(i, ql, min(block[ql]*T, qr)) { 52 int t = Query(ql, qr, a[i]); 53 if (t > maxx || (t == maxx && val[a[i]] < val[res])) { 54 maxx = t; 55 res = a[i]; 56 } 57 } 58 if (block[ql] != block[qr]) { 59 rep(i, (block[qr]-1) * T + 1, qr) { 60 int t = Query(ql, qr, a[i]); 61 if (t > maxx || (t == maxx && val[a[i]] < val[res])) { 62 maxx = t; 63 res = a[i]; 64 } 65 } 66 } 67 return val[res]; 68 } 69 70 int main() { 71 n = ri, m = ri; 72 T = 300; 73 74 rep(i, 1, n){ 75 a[i] = ri; 76 if (!mp[a[i]]) { 77 val[++id] = a[i]; 78 mp[a[i]] = id; 79 } 80 v[mp[a[i]]].push_back(i); 81 a[i] = mp[a[i]]; 82 } 83 rep(i, 1, n) block[i] = (i-1) / T + 1; 84 rep(i, 1, block[n]) pre(i); 85 86 while (m--) { 87 int ql = ri, qr = ri; 88 ql = (ql + last - 1) % n + 1; 89 qr = (qr + last - 1) % n + 1; 90 if (ql > qr) swap(ql, qr); 91 printf("%d ", last = Query(ql, qr)); 92 } 93 return 0; 94 }
CH#46A，分块下的排序处理便于降低之后bfs的复杂度。
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <cctype> 5 #include <iostream> 6 #include <algorithm> 7 #define ll long long 8 #define ri readint() 9 #define gc getchar() 10 #define rep(i, l, r) for (int i = l; i <= r; i++) 11 using namespace std; 12 13 const int N = 25 * 1e4 + 5; 14 struct Stone { 15 ll x, y, m, p, r, d; 16 }q[N], st[N]; 17 int L[N], R[N], M[N]; 18 int n, T, block; 19 bool mark[N]; 20 21 inline int readint() { 22 int x = 0, s = 1, c = gc; 23 while (c <= 32) c = gc; 24 if (c == '-') s = -1, c = gc; 25 for (; isdigit(c); c = gc) 26 x = x * 10 + c - 48; 27 return x * s; 28 } 29 30 inline ll sqr(ll x) { 31 return x * x; 32 } 33 34 inline bool cmp1(const Stone &a, const Stone &b) { 35 return a.m < b.m; 36 } 37 38 inline bool cmp2(const Stone &a, const Stone &b) { 39 return a.d < b.d; 40 } 41 42 void READ() { 43 q[0].x = ri, q[0].y = ri, q[0].p = ri, q[0].r = ri; 44 n = ri; 45 T = sqrt(n); 46 rep(i, 1, n) { 47 st[i].x = ri; 48 st[i].y = ri; 49 st[i].m = ri; 50 st[i].p = ri; 51 st[i].r = ri; 52 st[i].d = sqr(st[i].x - q[0].x) + sqr(st[i].y - q[0].y); 53 } 54 } 55 56 void BLOCK() { 57 sort(st+1, st+1+n, cmp1); 58 block = (n - 1) / T + 1; 59 rep(i, 1, block) { 60 L[i] = (i - 1) * T + 1; 61 R[i] = min(i * T, n); 62 M[i] = st[R[i]].m; 63 sort(st+L[i], st+R[i]+1, cmp2); 64 } 65 } 66 67 int BFS() { 68 int l = 0, r = 1; 69 while (l < r) { 70 int k = 0; 71 rep(i, 1, block) { 72 if (M[i] > q[l].p) 73 break; 74 k = i; 75 } 76 77 rep(i, 1, k) { 78 while (L[i] <= R[i] && st[L[i]].d <= sqr(q[l].r)) { 79 if (!mark[L[i]]) { 80 mark[L[i]] = true; 81 q[r++] = st[L[i]]; 82 } 83 ++L[i]; 84 } 85 } 86 if (block != k++) { 87 rep(i, L[k], R[k]) { 88 if (!mark[i] && st[i].m <= q[l].p && st[i].d <= sqr(q[l].r)) { 89 mark[i] = true; 90 q[r++] = st[i]; 91 } 92 } 93 } 94 95 ++l; 96 } 97 return r - 1; 98 } 99 100 int main() { 101 READ(); 102 BLOCK(); 103 cout << BFS() << endl; 104 return 0; 105 }
BZOJ2038，喜闻乐见的莫队袜子，感觉求概率这个数学式子的变换也是很重要的一环呀。然后分块排序保证了询问均摊根号N的复杂度吧。
1 #include <bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 5 const int N = 1e5 + 5; 6 int n, m, col[N]; 7 int T, block[N]; 8 ll ans, sum[N]; 9 struct Mo { 10 int l, r, id; 11 ll A, B; 12 }q[N]; 13 14 bool cmp1(Mo a, Mo b) { 15 return block[a.l] == block[b.l] ? a.r < b.r : a.l < b.l; 16 } 17 18 bool cmp2(Mo a, Mo b) { 19 return a.id < b.id; 20 } 21 22 ll sqr(ll x) { 23 return x * x; 24 } 25 26 void change(int x, int add) { 27 ans -= sqr(sum[col[x]]); 28 sum[col[x]] += add; 29 ans += sqr(sum[col[x]]); 30 } 31 32 int main() { 33 cin >> n >> m; 34 T = sqrt(n); 35 for (int i = 1; i <= n; i++) { 36 scanf("%d", &col[i]); 37 block[i] = (i - 1) / T + 1; 38 } 39 for (int i = 1; i <= m; i++) { 40 scanf("%d%d", &q[i].l, &q[i].r); 41 q[i].id = i; 42 } 43 sort(q+1, q+1+m, cmp1); 44 45 int l = 1, r = 0; 46 for (int i = 1; i <= m; i++) { 47 while (l < q[i].l) change(l++, -1); 48 while (l > q[i].l) change(--l, 1); 49 while (r < q[i].r) change(++r, 1); 50 while (r > q[i].r) change(r--, -1); 51 52 if (q[i].l == q[i].r) { 53 q[i].A = 0ll; 54 q[i].B = 1ll; 55 continue; 56 } 57 ll t = q[i].r - q[i].l + 1; 58 q[i].A = ans - t; 59 q[i].B = sqr(t) - t; 60 ll gcd = __gcd(q[i].A, q[i].B); 61 if (!gcd) q[i].B = 1ll; 62 else q[i].A /= gcd, q[i].B /= gcd; 63 } 64 65 sort(q+1, q+1+m, cmp2); 66 for (int i = 1; i <= m; i++) 67 printf("%lld/%lld ", q[i].A, q[i].B); 68 69 return 0; 70 }