hdu 5200 Trees（离线处理，排序）

Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1383    Accepted Submission(s): 447

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.
 

题意：连在一起的叫一个块，现在给出n颗树的高度，再给出q个询问

每个询问给出一个x，输出把高度<=x的树都砍掉之后剩下几个块

思路：因为询问后项可以由前项继续推导获得，所以可以考虑用离线。

把所有的树的高度和询问都保存下来，排序后进行处理即可

过程有点繁琐，要耐心。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 50050
struct Node
{
    int v,id;
} p[N],a[N];
int flag[N];
int ans[N];
bool cmp(Node a,Node b)
{
    return a.v<b.v;
}
int main()
{
    int n,q;
    while(~scanf("%d %d",&n,&q))
    {
        memset(flag,0,sizeof(flag));
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i].v);
            if(!a[i].v) flag[i]=1;
            a[i].id=i;
        }
        for(int i=1; i<=q; i++)
        {
            scanf("%d",&p[i].v);
            p[i].id=i;
        }
        a[0].v=0;
        int cnt=1,num=0;
        for(int i=1; i<=n; i++)
            if(!a[i-1].v&&a[i].v>0)
                num++;
        sort(a+1,a+1+n,cmp);
        sort(p+1,p+1+q,cmp);
        flag[0]=flag[n+1]=1;
        for(int i=1; i<=q; i++)
        {
            if(p[i].v)
            {
                while(cnt<=n&&a[cnt].v<=p[i].v)
                {
                    int id=a[cnt++].id;
                    if(flag[id]) continue;
                    flag[id]=1;
                    if(!flag[id-1]&&!flag[id+1]) num++;
                    if(flag[id-1]&&flag[id+1]) num--;
                }
            }
            ans[p[i].id]=num;
        }
        for(int i=1; i<=q; i++)
            printf("%d\n",ans[i]);
    }
    return 0;
}