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hdu6406 Taotao Picks Apples 多校第8场1010

IT知识apple多校taotao

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Problem Description There is an apple tree in front of Taotao‘s house. When autumn comes,  n  apples on the tree ripen, and Taotao will go to pick these apples.

When Taotao picks apples, Taotao scans these apples from the first one to the last one. If the current apple is the first apple, or it is strictly higher than the previously picked one, then Taotao will pick this apple; otherwise, he will not pick.

Given the heights of these apples  h1,h2,?,hn, you are required to answer some independent queries. Each query is two integers  p,q, which asks the number of apples Taotao would pick, if the height of the  p-th apple were  q  (instead of  hp). Can you answer all these queries?   【hdu6406 Taotao Picks Apples 多校第8场1010】 
Input The first line of input is a single line of integer  T  (1≤ T≤ 10), the number of test cases.

Each test case begins with a line of two integers  n,m  (1≤ n,m≤ 105), denoting the number of apples and the number of queries. It is then followed by a single line of  n  integers  h1,h2,?,hn  (1≤ hi≤ 109), denoting the heights of the apples. The next  m  lines give the queries. Each of these  m  lines contains two integers  p  (1≤ p≤ n)  and  q  (1≤ q≤ 109), as described in the problem statement.    
Output For each query, display the answer in a single line.    
Sample Input15 31 2 3 4 41 55 52 3   
Sample Output153HintFor the first query, the heights of the apples were 5, 2, 3, 4, 4, so Taotao would only pick the first apple. For the second query, the heights of the apples were 1, 2, 3, 4, 5, so Taotao would pick all these five apples.For the third query, the heights of the apples were 1, 3, 3, 4, 4, so Taotao would pick the first, the second and the fourth apples.      预处理在RMQ+二分，挺好的题目。 dp1,为前缀，dp2为后缀。  

1 #include < bits/stdc++.h> 2 using namespace std; 3 const int N = 1e5+10; 4 int dmax[N][25],a[N], cnt[N], dp1[N], dp2[N]; 5 int que[N], t, v, l, n, m; 6 void init(){ 7for(int i = 1; i < = n; i ++){ 8dmax[i][0] = a[i]; 9} 10for(int j = 1; (1< < j) < = n; j ++){ 11for(int i = 1; i+(1< < j)-1< = n; i ++){ 12dmax[i][j] = max(dmax[i][j-1],dmax[(1< < (j-1))+i][j-1]); 13} 14} 15 } 16 int getValue(int l, int r){ 17int k = 0; 18while(1< < (k+1) < = (r-l+1))k++; 19return max(dmax[l][k],dmax[r-(1< < k)+1][k]); 20 } 21 int find(int l, int r, int x) { 22int pos = -1; 23while(l < = r) { 24int m = (l+r) > > 1; 25if(getValue(l, m) > x) { 26pos = m; 27r = m-1; 28} else l = m+1; 29} 30return pos; 31 } 32 33 int main() { 34scanf("%d", & t); 35while(t--) { 36scanf("%d%d", & n, & m); 37int mx = 0; 38for(int i = 1; i < = n; i ++) { 39scanf("%d", & a[i]); 40if(mx < a[i]) { 41mx = a[i]; 42dp1[i] = dp1[i-1]+1; 43} else dp1[i] = dp1[i-1]; 44cnt[i] = max(a[i], cnt[i-1]); 45} 46init(); 47int tail = 0, head = 1; 48for(int i = n; i > = 1; i --) { 49while(head < = tail & & que[tail] < = a[i]) tail --; 50que[++tail] = a[i]; 51dp2[i] = (tail- head + 1); 52} 53while(m--) { 54scanf("%d%d", & l, & v); 55int ans = dp1[l-1]; 56if(v > cnt[l-1]) ++ ans; 57v = max(v, cnt[l-1]); 58int pos = find(l+1, n, v); 59if(pos != -1) ans += dp2[pos]; 60printf("%d ",ans); 61} 62} 63return 0; 64 }

 

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