HDU - 5200 - Trees （upper_bound）

Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 205    Accepted Submission(s): 68

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 \leq N, Q \leq 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 252362

 

Sample Output

02
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.
 

 

Source

BestCoder Round #36 ($)

 

STL中关于二分查找的函数有三个lower_bound 、upper_bound 、binary_search 。这三个函数都运用于有序区间（当然这也是运用二分查找的前提），下面记录一下这两个函数。

ForwardIter lower_bound(ForwardIter first, ForwardIter last,const _Tp& val)算法返回一个非递减序列[first, last)中的第一个大于等于值val的位置。

ForwardIter upper_bound(ForwardIter first, ForwardIter last, const _Tp& val)算法返回一个非递减序列[first, last)中的第一个大于值val的位置。

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn = 100005;int n, q;int h[maxn], ans[maxn], vis[maxn];pair<int, int> p[maxn];int main() {while(scanf("%d %d", &n, &q) != EOF) {for(int i = 1; i <= n; i++) {scanf("%d", h + i);p[i] = make_pair(-h[i], i);}memset(vis, 0, sizeof(vis));sort(p + 1, p + n + 1);//因为存的是负值，所以是从大到小排序 int sum = 0;//记录当前有多少个块 for(int i = 1; i <= n; i++) {int x = p[i].second;if(!vis[x - 1] && !vis[x + 1]) sum ++;//多一个块，因为两边都没被访问 else if(vis[x - 1] && vis[x + 1]) sum --;//合并两个块为一个，因为正好将两边连了起来，其他情况不会影响块的增加与减少  ans[i] = sum;//这里表示从第i个数据以下（数据是从大到小的）射击能够得到的块数vis[x] = 1;}for(int i = 0; i < q; i++) {int t;scanf("%d", &t);//upper_bound返回一个非递减序列[first, last)中的第一个大于值val的位置。 int id = upper_bound(p + 1, p + n + 1, make_pair(-t, -1)) - p - 1;printf("%d\n", ans[id]);}}return 0;}