H - Typesetting HDU - 6107

Yellowstar is writing an article that contains N words and 1 picture, and the i-th word contains aiaicharacters. 
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. 

InputThe 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 aiai, indicates i-th word consists of aiai 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≤10T≤10 
1≤N,W,Q≤1051≤N,W,Q≤105 
1≤pw,ai≤W1≤pw,ai≤W 
0≤dw≤W?pw0≤dw≤W?pwOutputFor 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

题解：　　这个题目，的确，第一眼看上去是在倍增什么呢？可以倍增字节，但是复杂度还是不对，那么还有什么量可以倍增呢？就只剩下行数了，好久没有写倍增了，什么边界处理什么的都忘了，所以讲一下实现吧。　　dp[i][j]表示在没有图的情况下以第i个单词为开头，覆盖2^j次方单词包括i在内需要几个，这个初始状态就是对于每个单词，暴力匹配他还需要多少单词，付给dp[i][0]，然后有一个显然的状态转移dp[i][j]=dp[i][j-1]+dp[i+dp[i][j-1]][j-1];这里处理一下边界，如果dp[i][j-1]||dp[i+dp[i][j-1]][j-1]==0就不处理，保证之后单词数不够的情况下，dp[i][j]=0。　　然后再次设一个状态dp2[i][j]表示在有图的情况下以第i个单词为开头，覆盖2^j次方单词包括i在内需要几个，初始值和状态转移都是一样的。求的时候因为不可能超过20，就从20for到0，能跳就跳就可以了。　　具体就是没到图的开头时用dp1跳到图的开头，然后用dp2，跳过图中，最后如果还剩下单词就继续用dp跳，有细节，推荐看一下代码。

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <iostream>
#define MAXN 101000
using namespace std;
int dp[MAXN][22],dp2[MAXN][22];
int hi[MAXN],star[MAXN];
int a[MAXN];
int n,w,pw,dw,q;

void init(){
    scanf("%d%d%d%d",&n,&w,&pw,&dw);
    memset(a,0,sizeof(a));
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    scanf("%d",&q);
    memset(star,0,sizeof(star));
    memset(hi,0,sizeof(hi));
    for(int i=1;i<=q;i++) scanf("%d%d",&star[i],&hi[i]);
}

void pre(){
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=n;i++){
        int sum=a[i],j=i+1;
    while(sum+a[j]+1<=w&&j<=n) sum+=a[j++]+1;
        dp[i][0]=j-i;
    }
    for(int j=1;j<=20;j++)
        for(int i=1;i<=n;i++){
            if(!dp[i][j-1]||!dp[i+dp[i][j-1]][j-1]) continue;
            dp[i][j]=dp[i][j-1]+dp[i+dp[i][j-1]][j-1];
        }
    memset(dp2,0,sizeof(dp2));
    for(int i=1;i<=n;i++){
        int j=i,flag=0,sum=0;
        while(sum+a[j]+flag<=dw&&j<=n) sum+=a[j++]+flag,flag=1;
        int k=j;
        sum=0,flag=0;
        while(sum+a[k]+flag<=w-pw-dw&&k<=n) sum+=a[k++]+flag,flag=1;
        dp2[i][0]=k-i;
    }
    for(int j=1;j<=20;j++)
        for(int i=1;i<=n;i++){
            if(!dp2[i][j-1]||!dp2[i+dp2[i][j-1]][j-1]) continue;
            dp2[i][j]=dp2[i][j-1]+dp2[i+dp2[i][j-1]][j-1];
        }
}

int jump1(int now){
    int ans=0;
    for(int j=20;j>=0;j--){
        if(!dp[now][j]) continue;
        if(now+dp[now][j]-1<=n){
            now+=dp[now][j];
            ans+=1<<j;
        }
    }
    return ans;
}

int jump2(int now,int cen){
    for(int j=20;j>=0;j--){
        if(!dp[now][j]) continue;
        if((1<<j)<=cen){
            cen-=(1<<j);
            now+=dp[now][j];
        }
    }
    return now;
}

int jump3(int now,int cen){
    for(int j=20;j>=0;j--){
        if(!dp2[now][j]) continue;
        if((1<<j)<=cen){
            cen-=(1<<j);
            now+=dp2[now][j];
        }
    }
    return now;
}

void work(){
    for(int i=1;i<=q;i++){
        int x=star[i],h=hi[i],tmp=jump1(1);
        if(tmp<=x-1){
            printf("%d\n",tmp+h);
            continue;
        }
        int ans=x+h-1,now=jump2(1,x-1);
        now=jump3(now,h);
        if(now<=n) ans+=jump1(now);
        printf("%d\n",ans);
    }
}

int main()
{
    int t;cin>>t;
    while(t--){
        init();
        pre();
        work();
    }
    return 0;
}