hdu 5200 Trees 排序+思维? ★★☆

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Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1282    Accepted Submission(s): 398


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]

1N,Q50000

0h[i]1000000000(109)

0q[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 ($)
 

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#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<climits>#include<queue>#include<vector>#include<map>#include<sstream>#include<set>#include<stack>#include<cctype>#include<utility>#pragma comment(linker, "/STACK:102400000,102400000")#define PI (4.0*atan(1.0))#define eps 1e-10#define sqr(x) ((x)*(x))#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)#define  lson   ind<<1,le,mid#define rson    ind<<1|1,mid+1,ri#define MID   int mid=(le+ri)>>1#define zero(x)((x>0? x:-x)<1e-15)#define mk    make_pair#define _f     first#define _s     secondusing namespace std;//const int INF=    ;typedef long long ll;//const ll inf =1000000000000000;//1e15;//ifstream fin("input.txt");//ofstream fout("output.txt");//fin.close();//fout.close();//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);const int INF =0x3f3f3f3f;const int maxn= 50000+20    ;//const int maxm=    ;  int n,q; int sum[maxn];bool vis[maxn];struct Node{   int hei,ind;  }a[maxn]; bool cmp(Node x,Node y){    return x.hei>y.hei;} int search(int x){    int le=1,ri=n;    while(le<=ri)    {        int mid=(le+ri)/2;        if(a[mid].hei<=x) ri=mid-1;        else     le=mid+1;    }    return ri;}int main(){    int x;  while(~scanf("%d%d",&n,&q))  {      FOR1(i,n)  scanf("%d",&a[i].hei),a[i].ind=i;      sort(a+1,a+1+n,cmp);       memset(vis,0,sizeof vis);      sum[0]=0,sum[1]=1;       vis[ a[1].ind ] =1;        for(int i=2;i<=n;i++)      {           x=a[i].ind;          if( (vis[x-1]&& !vis[x+1]) || (!vis[x-1]&& vis[x+1])  )          {              sum[i]=sum[i-1];          }          else if(  vis[x-1]&&vis[x+1]   )          {              sum[i]=sum[i-1]-1;          }          else          {              sum[i]=sum[i-1]+1;          }          vis[x]=1;      }      /*      for(int i=1;i<=n;i++)      {          cout<<i<<" "<<sum[i]<<endl;      }*/       while(q--)      {          scanf("%d",&x);          int p=search(x);          printf("%d\n",sum[p]);      }  }     return 0;}/*6  312 7 10 15 7 871012*/


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