hdu 6107--Typesetting(倍增)

Posted 2020-10-05 茶飘香~

题目链接

 

Problem Description

Yellowstar is writing an article that contains N words and 1 picture, and the i-th word contains ai characters.
The page width is fixed to W characters. In order to make the article look more beautiful, Yellowstar has made some rules:

1. The fixed width of the picture is pw. The distance from the left side of the page to the left side of the photo fixed to dw, in other words, the left margin is dw, and the right margin is W - pw - dw.
2. The photo and words can‘t overlap, but can exist in same line.
3. The relative order of words cannot be changed.
4. Individual words need to be placed in a line.
5. If two words are placed in a continuous position on the same line, then there is a space between them.
6. Minimize the number of rows occupied by the article according to the location and height of the image.

However, Yellowstar has not yet determined the location of the picture and the height of the picture, he would like to try Q different locations and different heights to get the best look. Yellowstar tries too many times, he wants to quickly know the number of rows each time, so he asked for your help. It should be noted that when a row contains characters or pictures, the line was considered to be occupied.

 

 

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with four integers N, W, pw, dw : the number of words, page width, picture width and left margin.
The next line contains N integers ai, indicates i-th word consists of ai characters.
The third line contains one integer Q. 
Then Q lines follow, each line contains the values of xi and hi, indicates the starting line and the image height of the image.

Limits
T≤10
1≤N,W,Q≤105
1≤pw,ai≤W
0≤dw≤W−pw

 

 

Output

For each query, output one integer denotes the minimum number of rows.

 

 

Sample Input

2

2 7 4 3

1 3

3

1 2

2 2

5 2

3 8 2 3

1 1 3

1

1 1

 

 

Sample Output

2

3

3

1

 

 

题意：现在有一篇文章由 n 个单词构成，其中要放入一张图片，单词与单词之间要空一格，图片上不能放单词，单词之间的顺序不能变化，现在规定图片宽为pw，文章宽为w，图片左边距文章左边线宽为dw，现在Q次询问，每次给定图片高h和图片在文章中的起始位置（距离文章上边界）x，求最少多少行能放下图片与单词？

 

思路：先计算出固定宽为w、dw、w-dw-pw时以每个单词开头时下一行第一个单词的下标，根据dw和w-dw-pw可以计算出有图片占据行时以每个单词开始时的下一行第一个单词的下标，然后根据已求出的有图片和无图片占据行时以每个单词开始的下一行第一个单词的下标信息，进行倍增求出以每个单词开始时，向下2^i行的第一个单词的下标。

 

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=1e5+5;
int n,w,pw,dw;
int a[N];
int lto[N],rto[N],to[N];
int s[N][20],t[N][20];

void cal(int *c,int W)
{
    int len=-1;
    for(int i=0,j=0;i<=n;i++)
    {
        while(len+a[j]+1<=W) len+=a[j]+1,j++;
        c[i]=j;
        len-=a[i]+1;
    }
}
void process()
{
    cal(lto,w);
    for(int i=0;i<=n;i++) s[i][0]=lto[i];
    to[n]=0;
    for(int i=n-1;i>=0;i--) to[i]=to[lto[i]]+1;
    for(int i=1;i<18;i++)
        for(int j=0;j<=n;j++)
           s[j][i]=s[s[j][i-1]][i-1];
    cal(lto,dw);
    cal(rto,w-dw-pw);
    for(int i=0;i<=n;i++) t[i][0]=rto[lto[i]];
    for(int i=1;i<18;i++)
        for(int j=0;j<=n;j++)
           t[j][i]=t[t[j][i-1]][i-1];
}
int main()
{
    int T; cin>>T;
    while(T--)
    {
       scanf("%d%d%d%d",&n,&w,&pw,&dw);
       for(int i=0;i<n;i++) scanf("%d",&a[i]);
       a[n]=w+5;
       process();
       int Q; scanf("%d",&Q);
       while(Q--)
       {
           int x,h,ans,tot=0; scanf("%d%d",&x,&h);
           ans=min(--x,to[0]);
           for(int i=0,tmp=ans;i<18;i++)
           {
               if(tmp&1) tot=s[tot][i];
               tmp>>=1;
           }
           for(int i=0,tmp=h;i<18;i++)
           {
               if(tmp&1) tot=t[tot][i];
               tmp>>=1;
           }
           ans+=h;
           ans+=to[tot];
           printf("%d\n",ans);
       }
    }
    return 0;
}