BC.5200.Trees(dp)

Trees

 Accepts: 156

 Submissions: 533

 Time Limit: 2000/1000 MS (Java/Others)

 Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Today CodeFamer is going to cut trees.There are N trees standing in a line. They are numbered from 1 to N. The tree numbered i has height hi. We say that two uncutted trees whose numbers are x and y are in the same block if and only if they are fitting in one of blow rules:

1)x+1=y or y+1=x;

2)there exists an uncutted tree which is numbered z, and x is in the same block with z, while y is also in the same block with z.

Now CodeFamer want to cut some trees whose height is not larger than some value, after those trees are cut, how many tree blocks are there?

Input

Multi test cases (about 15).

For each case, first line contains two integers N and Q separated by exactly one space, N indicates there are N trees, Q indicates there are Q queries.

In the following N lines, there will appear h[1],h[2],h[3],…,h[N] which indicates the height of the trees.

In the following Q lines, there will appear q[1],q[2],q[3],…,q[Q] which indicates CodeFamer’s queries.

Please process to the end of file.

[Technical Specification]

1≤N,Q≤50000

0≤h[i]≤1000000000(109)

0≤q[i]≤1000000000(109)

Output

For each q[i], output the number of tree block after CodeFamer cut the trees whose height are not larger than q[i].

Sample Input

3 2
5
2
3
6
2

Sample Output

0
2

Hint

In this test case, there are 3 trees whose heights are 5 2 3. For the query 6, if CodeFamer cuts the tree whose height is not large than 6, the height form of left trees are -1 -1 -1(-1 means this tree was cut). Thus there is 0 block. For the query 2, if CodeFamer cuts the tree whose height is not large than 2, the height form of left trees are 5 -1 3(-1 means this tree was cut). Thus there are 2 blocks.

 #include<stdio.h>
 #include<map>
 #include<algorithm>
 #include<set>
 #include<string.h>
 using namespace std;
 int n , m ;
 const int M =  +  ;
 int cut[M] ;
 bool vis[M] ;
 pair <int , int> a[M] ;

 void solve ()
 {
     for (int i =  ; i <= n ; i++) {
         scanf ("%d" , &a[i].first) ;
         a[i].second = i ;
     }
     sort (a +  , a + n + ) ;
     for (int i = n ; i >=  ; i --) {
         cut [i] = cut [i + ] ;
         int now = a[i].second ;
         if (vis[now - ] && vis[now + ] ) {
             cut[i] -- ;
         }
         else if (!vis[now - ] && !vis[now + ]) {
             cut[i] ++ ;
         }
         vis[now] =  ;//1 代表没砍 ， 0 代表砍了
     }
     while (m--) {
         int h ;
         scanf ("%d" , &h) ;
         int x = upper_bound (a +  , a + n +  , make_pair (h , n + ) ) - a ;
         if (x > n) puts ("") ; else printf ("%d\n" , cut[x]) ;
     }
 }

 int main ()
 {
    // freopen ("a.txt" , "r" , stdin ) ;
     while (~ scanf ("%d%d" , &n , &m)) {
         memset (vis ,  , sizeof(vis)) ;
         memset (cut ,  , sizeof(cut)) ;
         solve () ;
     }
     return  ;
 }

假设0为砍掉状态 ， 1 为活着状态。

假设一排树的高度为               2 4 3 1 6 5

他们一开始都为                      0 0 0 0 0 0

若只有最高的活着，状态变为  0 0 0 0 1 0   sta[6] = 1

若有第2高的活着，状态为       0 0 0 0 1 1   sta[5] = sta[6] = 1 , 显而易见第5，6棵树构成block , 不用+ 1

若有第3高的活者，状态为       0 1 0 0 1 1   sta[4] = sta[5] + 1 = 2 , 因为第{2} ， {5 ， 6}都为block

若有第4高的活着，状态为       0 1 1 0 1 1   sta[3] = sta[4]  = 2 ; {2 ， 3} ， {5 ， 6}

如有第5高的活着，状态为       1 1 1 0 1 1　sta[2] = sta[3] = 2 ;  {1 , 2 , 3} , {5 , 6}

如果都活着， 状态为               1 1 1 1 1 1   sta[1] = sta[2] - 1 ; 只有{1 ， 2 ，3 ， 4 ， 5 ， 6} 一个block

虽然很orz，但过然是dp吧