ZOJ-3635 Cinema in Akiba (tree array + dichotomy)

Test instructions: Known to have n individuals, starting from the first person each person is arranged on the AI empty seat, there is M group inquiry, ask someone to sit position.

Analysis:

1, the tree-like array to maintain the number of empty seats, method:

Initializing all empty seats to 1,sum (x) indicates the number of seats from seat 1 to seat x empty seat.

2, for each person, according to sum (mid), two points for sum (mid) greater than or equal to A[i] the smallest mid, that is, the position of the AI empty block, and add 1 to that position, then the value of that position becomes 0, thus not participating in the statistics of the number of empty seats.

3, Vis[q] is the position of the person who is labeled Q.

#include <cstdio> #include <cstring> #include <cstdlib> #include <cctype> #include <cmath > #include <iostream> #include <sstream> #include <iterator> #include <algorithm> #include <string> #include <vector> #include <set> #include <map> #include <stack> #include < deque> #include <queue> #include <list> #define LOWBIT (x) (X & (x)) const DOUBLE EPS = 1e-8;inline int    DCMP (double A, double b) {if (Fabs (a) < EPS) return 0; Return a > B? 1:-1;} typedef long Long Ll;typedef unsigned long long ull;const int int_inf = 0x3f3f3f3f;const int int_m_inf = 0x7f7f7f7f;const ll ll_inf = 0x3f3f3f3f3f3f3f3f;const LL ll_m_inf = 0x7f7f7f7f7f7f7f7f;const int dr[] = {0, 0,-1, 1,-1,-1, 1, 1};const I NT Dc[] = {-1, 1, 0, 0,-1, 1,-1, 1};const int MOD = 1e9 + 7;const Double pi = ACOs ( -1.0); const int MAXN = 50000 + 10;con St int MAXT = 10000 + 10;using namespace Std;int vis[maxn];int a[maxn];int z[maxn];int n;int sum (int x) {int ans = 0;    for (int i = x; I >= 1; I-= Lowbit (i)) {ans + = z[i]; } return ans;    void Add (int x, int value) {for (int i = x; i <= n; i + = Lowbit (i)) {Z[i] + = value;    }}int solve (int x) {int L = 1, r = N;        while (L < r) {int mid = L + (r-l)/2;        if (sum (mid) >= x) r = Mid;    else L = mid + 1; } return R;}        int main () {while (scanf ("%d", &n) = = 1) {memset (Vis, 0, sizeof vis);            for (int i = 1; I <= n; ++i) {scanf ("%d", &a[i]);        Add (i, 1);            } for (int i = 1; I <= n; ++i) {vis[i] = solve (A[i]);        Add (Vis[i],-1);        } int m;        scanf ("%d", &m);        BOOL flag = TRUE;            while (m--) {int q;            scanf ("%d", &q);            if (flag) flag = false;            else printf ("");        printf ("%d", vis[q]);    } printf ("\ n"); } return 0;}

　　

ZOJ-3635 Cinema in Akiba (tree array + dichotomy)