Super Mario
Problem Description
Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.
Input
The first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)
Output
For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.
Sample Input
1
10 10
0 5 2 7 5 4 3 8 7 7
2 8 6
3 5 0
1 3 1
1 9 4
0 1 0
3 5 5
5 5 1
4 6 3
1 5 7
5 7 3
Sample Output
Case 1:
4
0
0
3
1
2
0
1
5
1
Source
2012 ACM/ICPC Asia Regional Hangzhou Online
题目类型：划分树+二分枚举
算法分析：二分枚举每次要查询的区间长度(即rank)，之后就是简单的判断
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/************************************************* Author :supermaker Created Time :2016/1/15 0:17:04 File Location :C:\Users\abcd\Desktop\TheEternalPoet **************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e5 + 6666; int sorted[maxn], tree[30][maxn], toleft[30][maxn]; void build (int dep, int l, int r) { if (l == r) return ; int mid = (l + r) >> 1; int same = mid - l + 1; for (int i = l; i <= r; i++) if (tree[dep][i] < sorted[mid]) same--; int lpos = l, rpos = mid + 1; for (int i = l; i <= r; i++) { if (tree[dep][i] < sorted[mid]) tree[dep+1][lpos++] = tree[dep][i]; else if (tree[dep][i] == sorted[mid] && same > 0) { tree[dep+1][lpos++] = tree[dep][i]; same--; } else tree[dep+1][rpos++] = tree[dep][i]; toleft[dep][i] = toleft[dep][l-1] + (lpos - l); } build (dep + 1, l, mid); build (dep + 1, mid + 1, r); } int query (int dep, int L, int R, int l, int r, int k) { if (l == r) return tree[dep][l]; int mid = (L + R) >> 1; int tot = toleft[dep][r] - toleft[dep][l-1]; if (tot >= k) { int newl = L + toleft[dep][l-1] - toleft[dep][L-1]; int newr = newl + tot - 1; return query (dep + 1, L, mid, newl, newr, k); } else { int newr = r + toleft[dep][R] - toleft[dep][r]; int newl = newr - (r - l - tot); return query (dep + 1, mid + 1, R, newl, newr, k - tot); } } int main() { //CFF; //CPPFF; int t, flag = 1; scanf ("%d", &t); while (t--) { int n, q; memset (tree, 0, sizeof (tree)); memset (toleft, 0, sizeof (toleft)); scanf ("%d%d", &n, &q); for (int i = 1; i <= n; i++) { scanf ("%d", &tree[0][i]); sorted[i] = tree[0][i]; } sort (sorted + 1, sorted + 1 + n); build (0, 1, n); printf ("Case %d:\n", flag++); for (int i = 1; i <= q; i++) { int ll, rr, kk; scanf ("%d%d%d", &ll, &rr, &kk); ll++, rr++; int low = 1, high = rr - ll + 1, res = 0; while (low <= high) { int mid = (low + high) >> 1; int val = query (0, 1, n, ll, rr, mid); if (val <= kk) low = mid + 1, res = mid; else high = mid - 1; } printf ("%d\n", res); } } return 0; } |
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