傳送門(權限題)

題目分析

題意為：給定一個數列，修改和查詢兩種操作，修改每次給定一個區間，區間的所有元素都加上一個給定值，查詢詢問一段區間的數權值大于等于給定值的數有多少個。 首先對原序列分塊，然后將分塊后的數組每個塊內的數字進行排序，這樣查詢時就可以暴力枚舉散塊，并二分枚舉每個整塊。對于修改，對于整塊部分只需修改tag，然后暴力修改散塊的原序列值，然后對散塊元素所在塊進行重排序即可。

code

3048?ms

#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> #include<vector> using namespace std;const int N = 1e6 + 5; int n, Q, h[N], H[N], tag[N], S, blo; #define bug(x) cout<<#x<<":"<<x<<endl inline int read(){int i = 0, f = 1; char ch = getchar();for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());if(ch == '-') f = -1, ch = getchar();for(; ch >= '0' && ch <= '9'; ch = getchar())i = (i << 3) + (i << 1) + (ch - '0');return i * f; }inline void wr(int x){if(x < 0) putchar('-'),x = -x;if(x > 9) wr(x / 10);putchar(x % 10 + '0'); }inline void add(int x, int y, int v){if(y - x + 1 <= 2 * S){for(int i = x; i <= y; i++)h[i] += v;}int Bx = x / S + (x % S ? 1 : 0), By = y / S + (y % S ? 1 : 0);int L = Bx + 1, R = By - 1;if(x == (Bx - 1) * S + 1) L--;if(y == min(n, By * S)) R++;int ans = 0;for(int i = x; i < (L - 1) * S; i++)h[i] += v;int l = (Bx - 1) * S + 1, r = min(n, Bx * S);for(int i = l; i <= r; i++)H[i] = h[i];sort(H + l, H + r + 1);for(int i = min(n, R * S) + 1; i <= y; i++)h[i] += v;l = (By - 1) * S + 1, r = min(n, By * S);for(int i = l; i <= r; i++)H[i] = h[i];sort(H + l, H + r + 1);for(int i = L; i <= R; i++)tag[i] += v; }inline int query(int x, int y, int v){int ans = 0;if(y - x + 1 <= 2 * S){for(int i = x; i <= y; i++)if(h[i] + tag[i / S + (i % S ? 1 : 0)] >= v) ans++;return ans;}int Bx = x / S + (x % S ? 1 : 0), By = y / S + (y % S ? 1 : 0); // bug(Bx), bug(By);int L = Bx + 1, R = By - 1;if(x == (Bx - 1) * S + 1) L--;if(y == min(n, By * S)) R++; // bug(L), bug(R);for(int i = x; i <= (L - 1) * S; i++)if(h[i] + tag[Bx] >= v) ans++;for(int i = min(n, R * S) + 1; i <= y; i++)if(h[i] + tag[By] >= v) ans++;for(int i = L; i <= R; i++){int l = (i - 1) * S + 1, r = min(n, i * S), len = r - l + 1;int tmp = lower_bound(H + l, H + r + 1, v - tag[i]) - (H + l - 1);ans += len - (tmp - 1);}return ans; }int main(){n = read(), Q = read(), S = sqrt(n), blo = n / S + (n % S ? 1 : 0); // bug(S);for(int i = 1; i <= n; i++) h[i] = H[i] = read();for(int i = 1; i <= blo; i++){int l = (i - 1) * S + 1, r = min(i * S, n);sort(H + l, H + r + 1);}for(int i = 1; i <= Q; i++){char opt[5]; scanf("%s", opt + 1);if(opt[1] == 'M'){int l = read(), r = read(), w = read();add(l, r, w);}else if(opt[1] == 'A'){int l = read(), r = read(), c = read();wr(query(l, r, c)), putchar('\n');}}return 0; }

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轉載于:https://www.cnblogs.com/CzYoL/p/7396352.html

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