"BZOJ4378" [Poi2015]logistyka "tree-like array" "concluding question"

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Exercises

First, there is a conclusion: the number of numbers greater than S is K (if K is greater than C, obviously Tak.) Here we discuss K <= c), then there must be a solution if the number is less than s and not less than (c-k) * S.

Will not prove that ...

Use a tree-like array to record the number of numbers and numbers, and then judge it.

Complexity of:

Complexity of Time: O (MLOGM), spatial complexity: O (M).

1 A.

GET:

Conclusion question ...

/* telekinetic Forest Guard */#include <cstdio> #include <cstring> #include <algorithm>using namespace Std;typedef Long Long ll;const int maxn = 1000005;int N, m, NUM[MAXN], DISC[MAXN], tot, TRNUM[MAXN]; LL trsum[maxn];struct _que {int opt, a, b;} que[maxn];inline int iread () {int f = 1, x = 0; char ch = getchar (); for (; Ch & Lt ' 0 ' | | Ch > ' 9 '; ch = getchar ()) F = ch = = '-'? -1:1;for (; Ch >= ' 0 ' && ch <= ' 9 '; ch = getchar ()) x = x * + ch-' 0 '; return f * x;} inline int find (int x) {int L = 1, R = Tot;while (L <= r) {int mid = L + R >> 1;if (Disc[mid] < x) L = mid + 1;e LSE if (Disc[mid] = = x) return mid;else r = mid-1;}} inline void Addnum (int x, int c) {for (; x <= tot; x + = x & x) trnum[x] + = C;} inline int querynum (int x) {int res = 0;for (; x; x = x & x) Res + = Trnum[x];return res;} inline void addsum (int x, int c) {for (; x <= tot; x + = x & x) trsum[x] + = C;} inline LL querysum (int x) {LL res = 0;for (; x; x-= x &-X) Res + = Trsum[x];return res;} int main () {iread (); m = Iread (); for (int i = 1, i <= m; i++) {Char ch = getchar (); for (; ch! = ' U ' && ch! = ' Z ') ; ch = getchar ()); que[i].a = Iread (); que[i].b = Disc[++tot] = Iread (); que[i].opt = (ch = = ' U ');} Sort (disc + 1, disc + 1 + tot), tot = unique (disc + 1, disc + 1 + tot)-(disc + 1); for (int i = 1; I <= m; i++) if (Que[i] . opt = = 1) {int x = que[i].a, y = Find (que[i].b), if (!num[x] && que[i].b) addnum (y, 1), N++;else if (Num[x] &&am p;!que[i].b) Addnum (y,-1), N--;else if (num[x] && que[i].b) addnum (Find (Num[x]),-1), Addnum (y, 1); if (Num[x]) add Sum (Find (num[x]),-num[x]), if (que[i].b) addsum (y, que[i].b); num[x] = que[i].b;} else {int x = que[i].a, y = Find (que[i].b), int k = N-querynum (y), if (k > x) printf ("tak\n"); else {LL sum = querysum (y) ;p rintf (sum >= (LL) (x-k) * que[i].b? "tak\n": "nie\n");}} return 0;}

"BZOJ4378" [Poi2015]logistyka "tree-like array" "concluding question"