HDU 5200 Trees

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 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.
按照高度排个序，从小到大取走这些树，根据条件判断剩下的块数。
然后读入数字后二分查找其位置，输出答案即可。
也可以选择读入询问并排序然后在计算块数时直接记录答案在按照标号输出。
#include<stdio.h>#include<algorithm>#include<map>#include<cstdlib>#include<cstring>using namespace std;map<int, int > M;const int maxn = 50005;int n, m, x, sum[maxn], f[maxn];struct abc{    int h, id;    abc(){}    abc(int h, int id) :h(h), id(id){}}a[maxn];bool cmp(const abc&a, const abc&b){    return a.h < b.h;}bool operator <(const int &a, const abc &b){    return a< b.h;}int main(){    while (scanf("%d%d", &n,&m)!=EOF)    {        memset(f, 0, sizeof(f));        memset(sum, 0, sizeof(sum));        for (int i = 0; i < n; i++)        {            scanf("%d", &x);            a[i] = abc(x, i);        }        sort(a, a + n, cmp);        for (int i = 0; i < n; i++)        {            if (i) sum[i] = sum[i - 1]; else sum[i] = 1;            if (a[i].id - 1 >= 0 && !f[a[i].id - 1] && a[i].id + 1 < n && !f[a[i].id + 1]) sum[i]++;            if ((a[i].id - 1 < 0 || f[a[i].id - 1]) && (a[i].id + 1 >= n || f[a[i].id + 1])) sum[i]--;            f[a[i].id] = 1;        }        while (m--)        {            scanf("%d", &x);            int k = upper_bound(a, a + n, x) - a;            if (x < a[k].h) k--;            if (k>=0) printf("%d\n", sum[k]);            else printf("1\n");        }    }}