Super Mario

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4046 Accepted Submission(s): 1866

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

题意：超级玛丽游戏的问题，很有意思。给你n个金币的高度（区间从0 - n-1）和m次查询，每次查询询问[l, r]里面小于或者等于h的金币数目。

思路：和树剖某一道题很像，sort下。把限制下的金币放里面，用个树状数组维护就好了。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (100000+10)
#define MAXM (300000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
int C[MAXN], n;
int lowbit(int x){return x & (-x);
}
void Update(int x, int d)
{while(x <= n){C[x] += d;x += lowbit(x);}
}
int Sum(int x)
{int s = 0;while(x > 0){s += C[x];x -= lowbit(x);}return s;
}
typedef pair<int, int> pii;
bool cmp1(pii a, pii b){return a.fi < b.fi;
}
vector<pii> G;
vector<pii> :: iterator it;
struct Node{int l, r, h, id;
};
Node In[MAXN];
bool cmp2(Node a, Node b){return a.h < b.h;
}
int ans[MAXN];
int main()
{int t, kcase = 1; Ri(t);W(t){int m; Ri(n); Ri(m); G.clear();for(int i = 1; i <= n; i++){int a; Ri(a);G.push_back(pii(a, i));}sort(G.begin(), G.end(), cmp1);for(int i = 1; i <= m; i++)Ri(In[i].l), Ri(In[i].r), Ri(In[i].h), In[i].id = i, In[i].l++, In[i].r++;sort(In+1, In+m+1, cmp2); CLR(C, 0); it = G.begin();for(int i = 1; i <= m; i++){while(it != G.end() && it->fi <= In[i].h) {Update(it->se, 1); it++;}ans[In[i].id] = Sum(In[i].r) - Sum(In[i].l-1);}printf("Case %d:\n", kcase++);for(int i = 1; i <= m; i++) Pi(ans[i]);}return 0;
}

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hdoj 4417 Super Mario 【树状数组 + 思维】

Super Mario

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