Super Mario

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12323 Accepted Submission(s): 5109

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

题目大意：输入N个数， M次询问，每次询问 L 到 R 之间 ≤ K 的数有多少个

思路：主席树是求某个区间第k小的数是哪个，这道题反过来了，告诉你数让你求第几小，所以可以二分答案，注意二分答案的时候如果一直等于的话也要一直更新，过程中存一下该数离散化之前是哪个数

Accepted code

#include<bits/stdc++.h>
#include<unordered_map>
using namespace std;#define sc scanf
#define ls rt << 1
#define rs ls | 1
#define Min(x, y) x = min(x, y)
#define Max(x, y) x = max(x, y)
#define ALL(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define MEM(x, b) memset(x, b, sizeof(x))
#define lowbit(x) ((x) & (-x))
#define P2(x) ((x) * (x))typedef long long ll;
const int MOD = 1e9 + 7;
const int MAXN = 1e5 + 100;
const int INF = 0x3f3f3f3f;
inline ll fpow(ll a, ll b){ ll r = 1, t = a; while (b){ if (b & 1)r = (r*t) % MOD; b >>= 1; t = (t*t) % MOD; }return r; }struct node
{int l, r, ans;
}t[MAXN * 20];
int p[MAXN], v[MAXN], n, m;
int root[MAXN], cnt, T, X;
vector <int> e;
void init() {MEM(t, 0); MEM(root, 0); cnt = 0;e.clear();
}
void Update(int l, int r, int &x, int y, int pos) {t[++cnt] = t[y]; t[cnt].ans++; x = cnt;if (l == r) return;int mid = (l + r) >> 1;if (mid >= pos) Update(l, mid, t[x].l, t[y].l, pos);else Update(mid + 1, r, t[x].r, t[y].r, pos);
}
int Query(int l, int r, int x, int y, int pos) {if (l == r) return l;int mid = (l + r) >> 1;int sum = t[t[y].l].ans - t[t[x].l].ans;if (sum >= pos) return Query(l, mid, t[x].l, t[y].l, pos);else return Query(mid + 1, r, t[x].r, t[y].r, pos - sum);
}
int Fun(int L, int R, int K) {int len = R - L + 1;int l = 1, r = len, scnt = 0;int max_ = Query(1, n, root[L - 1], root[R], len);  // 区间最大值int min_ = Query(1, n, root[L - 1], root[R], 1);  // 区间最小值if (v[max_] <= K) return len;   // 大于最大直接退出else if (v[min_] > K) return 0;  // 小于最小直接退出while (l <= r) {   // 二分答案int mid = (l + r) >> 1;int stot = Query(1, n, root[L - 1], root[R], mid); if (v[stot] <= K) scnt = mid, l = mid + 1;  // 一直等于一直更新else r = mid - 1; }return scnt;
}int main()
{cin >> T;while (T--) {sc("%d %d", &n, &m); init();for (int i = 1; i <= n; i++) {sc("%d", &p[i]);e.push_back(p[i]);}sort(ALL(e));e.erase(unique(ALL(e)), e.end());for (int i = 1; i <= n; i++) {int tot = p[i];p[i] = lower_bound(ALL(e), p[i]) - e.begin() + 1;Update(1, n, root[i], root[i - 1], p[i]);v[p[i]] = tot;  // 离散化之前的数}printf("Case %d:\n", ++X);for (int i = 0; i < m; i++) {int L, R, K;sc("%d %d %d", &L, &R, &K);printf("%d\n", Fun(L + 1, R + 1, K));}}return 0;
}

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#HDU 4417 Super Mario (主席树 + 二分)

Super Mario

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