简单搜索练习（一）二分入门 HDU 2141 Can you find it？POJ 2785 4 Value whose Sum is 0

简单搜索练习（一）：二分入门

最近事情比较多，忙各类社团的事情。再加上班里的一些事务，算法一直没怎么复习。一直想复习一下动态规划也就只看了一下书，前几天全校的排位赛挂机50min+心态差点崩掉，线段树、概率DP啥的一脸懵逼。学长一直给我们拉题做，说学好动态规划要先练好搜索。所以就一直开搜索专题。

一、二分搜索入门：

PS：二分的题目常见的就一般是二分查找存在性与二分答案两种

1、 HDU 2141 Can you find it？

Time Limit: 10000/3000 MS (Java/Others)

Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 33365

Accepted Submission(s): 8282

Problem Description

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.

Input

There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.

Output

For each case, firstly you have to print the case number as the form “Case d:”, then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print “YES”, otherwise print “NO”.

Sample Input

3 3 3

1 2 3

1 2 3

1 2 3

3

1

4

10

Sample Output

Case 1:

NO

YES

NO

题意

给你三个数组A，B，C，每个数组中含有N个数字，有M次询问，

每次给定一个X，问是否存在Ai+Bi+Ci=X。

Solution

看到题目估算一下暴力的时间复杂度500*500*500*1000，超过1e10，估计暴力过不了。一般来说，搜索的第一个优化就是二分。思路如下，先求出Ai+Bi然后sort一下，对于每一个X，二分查找X-Ci。

另一种方法是hash，本人太弱，把控不好

Attention

对于初学二分的人来说，二分的终止条件是最容易弄错的。有的时候需要进行一定的调整，不能死记模板。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int a[503],c[250003];
//二分查找 t 这里的二分查找条件有很多种
bool fi(int t,int l,int r)
{
int mid;
while(l<r-1)//注意终止条件
{
mid=(l+r)/2;
if(c[mid]<=t)
l=mid;
else
r=mid;
}
//不同的终止条件有不同的对应
if(c[l]==t||c[r]==t)
return true;
else
return false;
}
int main()
{
int t,k=0,size,l,n,m,b,size1,s;
bool su;
while(scanf("%d%d%d",&l,&n,&m)!=EOF)
{
printf("Case %d:\n",++k);
size=0;
for(int i=1;i<=l;i++)
scanf("%d",&a[i]);
//将A、B合并为一个
for(int i=1;i<=n;i++)
{
scanf("%d",&b);
for(int j=1;j<=l;j++)
c[++size]=b+a[j];
}
sort(c+1,c+1+size);
for(int i=1;i<=m;i++)
scanf("%d",&a[i]);
scanf("%d",&s);
for(int i=1;i<=s;i++)
{
su=false;//判断是否查找成功
scanf("%d",&b);
for(int j=1;j<=m;j++)
if(fi(b-a[j],1,size))
{
printf("YES\n");
su=true;
break;
}
if(!su)
printf("NO\n");
}
}
return 0;
}

PS:对于二分查找来说最重要的是要有序，sort非常关键。

2、POJ 2785 4 Values whose Sum is 0

Time Limit: 15000MS

Memory Limit: 228000K

Accepted: 7472

Case Time Limit: 5000MS

Description

The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

Input

The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 228 ) that belong respectively to A, B, C and D .

Output

For each input file, your program has to write the number quadruplets whose sum is zero.

Sample Input

6

-45 22 42 -16

-41 -27 56 30

-36 53 -37 77

-36 30 -75 -46

26 -38 -10 62

-32 -54 -6 45

Sample Output

5

Hint

Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

Source

Southwestern Europe 2005

题意

将上一题的三个数组变成了四个数组。上一个题不太会的可以那这个来练一下手。

Solution:

分别将A、B与C、D分别合成两个大数组，枚举其中一个再二分查找。本题的终止条件与上题不同。因为要找一共多少组，所以不要只找到一组就停止。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

long a[4003],b[4003],c[4003],d[4003];
long a1[16000003],a2[16000003];
long size,sum;
//二分查找开始
void fi(long  s)
{
long  l=1,r=size,mid;
while(l<r)//条件改变，如果用原来的条件会出错，卡掉部分解
{
mid=(l+r)/2;
if(a2[mid]+s<0)
l=mid+1;//边界的更改也相应改变
else
r=mid;
}
//查找相同的解有多少个，因为优先查找的比较小的所以l++
while(a2[l]+s==0&&l<=size)
{
sum++;
l++;
}
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
size=0;
sum=0;
for(int i=1;i<=n;i++)
scanf("%ld%ld%ld%ld",&a[i],&b[i],&c[i],&d[i]);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
a1[++size]=a[i]+b[j],a2[size]=c[i]+d[j];
sort(a1+1,a1+1+size);//有序化a1
sort(a2+1,a2+1+size);//有序化a2
for(int j=1;j<=size;j++)
fi(a1[j]);
printf("%ld\n",sum);
}
return 0;
}