SRM DIV2 586 PiecewiseLinearFunctionDiv2

【题意】根据Y画出一条折线，分别求出query中的直线与这样折线的交点数，并返回。

【算法】

1.求query[i]与折线中每天线段的交点数(端点不算)a,如果与某条线段重合，return-1,否则继续2;

2.求query[i]与Y中顶点的交点数b;

3.return a+b;

【调试bug】

1.17行判断query[i]与线段(Y[j],Y[j+1])是否有交点时，注意数字比较大，不能采用一般的乘法来判断。

【Java代码】来自菜鸟

import java.util.*;
import java.util.regex.*;
import java.text.*;
import java.math.*;


public class PiecewiseLinearFunctionDiv2
{
    public int[] countSolutions(int[] Y, int[] query)
    {
        int[] r = new int[query.length];
        int i,j;
        Arrays.fill(r, 0);
        
        for(i=0;i<query.length;i++){
            for(j=0;j<Y.length-1;j++){
                if(((query[i]-Y[j])+(query[i]-Y[j+1]))<Math.max(query[i]-Y[j], query[i]-Y[j+1])&&((query[i]-Y[j])+(query[i]-Y[j+1]))>Math.min(query[i]-Y[j], query[i]-Y[j+1]))
                    r[i]++;
                else if(query[i]==Y[j]&&query[i]==Y[j+1]){
                    r[i]=-1;
                    break;
                }
            }
            if(j<Y.length-1)
                continue;
            for(j=0;j<Y.length;j++){
                if(query[i]==Y[j])
                    r[i]++;
            }
        }
        return r;
    }
    

}
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【改进版代码】

【分析】判断query[i]与线段有没有交点的方法，更新了一下。

import java.util.*;
import java.util.regex.*;
import java.text.*;
import java.math.*;


public class PiecewiseLinearFunctionDiv2
{
    public int[] countSolutions(int[] Y, int[] query)
    {
        int[] r = new int[query.length];
        int i,j;
        Arrays.fill(r, 0);
        
        for(i=0;i<query.length;i++){
            for(j=0;j<Y.length-1;j++){
                if(query[i]>Y[j]&&query[i]<Y[j+1]||query[i]<Y[j]&&query[i]>Y[j+1])
                    r[i]++;
                else if(query[i]==Y[j]&&query[i]==Y[j+1]){
                    r[i]=-1;
                    break;
                }
            }
            if(j<Y.length-1)
                continue;
            for(j=0;j<Y.length;j++){
                if(query[i]==Y[j])
                    r[i]++;
            }
        }
        return r;
    }
    

}
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【Java代码】来自大神

【分析】算法差不多，只是在顶点的处理上有点不一样。

public class PiecewiseLinearFunctionDiv2 
{ 
  public static int[] countSolutions(int[] Y, int[] query) 
  { 
    int[] count = new int[query.length]; 
    for (int j = 0; j < query.length; j++) 
      if (Y[0] == query[j]) 
        count[j] = 1; 
    for (int i = 1; i < Y.length; i++) 
    { 
      for (int j = 0; j < query.length; j++) 
      { 
        if (count[j] == -1) 
          continue; 
        if (Y[i - 1] == query[j] && Y[i] == query[j]) 
          count[j] = -1; 
        if (Y[i - 1] > query[j] && Y[i] <= query[j] || Y[i - 1] < query[j] && Y[i] >= query[j]) 
          count[j]++; 
      } 
    } 
    return count; 
  } 
}

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【C++代码】来自大神

【分析】算法几乎一样的。

include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
 
using namespace std;
#define pb push_back
 
class PiecewiseLinearFunctionDiv2 {
public:
  vector <int> countSolutions(vector <int> Y, vector <int> query) {
    vector<int> ans;
    for(int i=0; i<query.size(); i++)
    {
      int temp=query[i];
      int cnt=0;
      int flag=0;
 
      for(int j=0; j<Y.size()-1; j++)
      {
        int y1=Y[j], y2=Y[j+1];
        if(temp==y1 && temp==y2)
          flag=1;
 
        if(temp>y1 && temp<y2)
          cnt++;
        else if(temp<y1 && temp>y2)
          cnt++;
      }
      for(int j=0; j<Y.size(); j++)
        if(Y[j]==temp)
          cnt++;
      if(flag==0)
        ans.pb(cnt);
      else
        ans.pb(-1);
    }
    return ans;
  }
};
 
 
 
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【总结】：题有点简单。