hdoj 2141 Can you find it? 【二分好题】【求STL set 怎么破】

最新推荐文章于 2016-10-09 12:24:00 发布

原创最新推荐文章于 2016-10-09 12:24:00 发布 · 522 阅读

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这篇博客介绍了如何通过给定三个数列A、B、C，找到符合条件的三个数Ai、Bj、Ck，使得其之和等于给定的数值X。通过将两个数列压缩成新数列T，并枚举第三个数列的元素来验证公式是否成立。

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Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 16926 Accepted Submission(s): 4305

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

Sample Output

Case 1:
NO
YES
NO

题意就不说了吧。。。

思路：把两个数组压缩成一个新数组T，枚举另外一个数组元素Y。对于要验证的结果X：则问题就变成了在数组T里面查找是否存在元素X - Y。

AC代码：561ms 在压缩新数组的时候，可以标记下某个元素是否存在，若存在不再进数组。在遍历另外一个数组元素时也可以标记下，这样应该能稍稍优化下吧。有兴趣的可以写下。

注意后面两个代码是错误的！！！

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 500+1
using namespace std;
int a[MAXN], b[MAXN], c[MAXN];
int d[250000+100];
int main()
{
    int L, N, M, S, X;
    int k = 1;
    int top;
    bool flag;
    while(scanf("%d%d%d", &L, &N, &M) != EOF)
    {
        for(int i = 0; i < L; i++)
            scanf("%d", &a[i]);
        for(int i = 0; i < N; i++)
            scanf("%d", &b[i]);
        for(int i = 0; i < M; i++)
            scanf("%d", &c[i]);
        //sort(a, a+L);
        //sort(b, b+N);
        top = 0;
        for(int i = 0; i < L; i++)
        {
            for(int j = 0; j < N; j++)
                d[top++] = a[i] + b[j];
        }
        sort(d, d+top);
        scanf("%d", &S);
        printf("Case %d:\n", k++);
        while(S--)
        {
            scanf("%d", &X);
            flag = false;
            for(int i = 0; i < M; i++)
            {
                int ans = X - c[i];//我们要搜索的结果
                int left = 0, right = top - 1, mid;
                while(right >= left)
                {
                    mid = (left + right) >> 1;
                    if(d[mid] == ans)
                    {
                        flag = true;
                        break;
                    }
                    else if(d[mid] > ans) right = mid - 1;
                    else left = mid + 1;
                }
                if(flag)
                    break;
            }
            if(!flag)
                printf("NO\n");
            else
                printf("YES\n");
        }
    }
    return 0;
}

错误代码一：

考虑到可能会有重复的元素，想用set优化下
用set存储新数组元素一直MLE，无语。。。。

#include <cstdio>
#include <cstring>
#include <set>
#include <algorithm>
#define MAXN 500+1
using namespace std;
int a[MAXN], b[MAXN], c[MAXN];
int main()
{
    int L, N, M, S, X;
    int k = 1;
    int top;
    bool flag;
    while(scanf("%d%d%d", &L, &N, &M) != EOF)
    {
        for(int i = 0; i < L; i++)
            scanf("%d", &a[i]);
        for(int i = 0; i < N; i++)
            scanf("%d", &b[i]);
        for(int i = 0; i < M; i++)
            scanf("%d", &c[i]);
        set<int> d;
        for(int i = 0; i < L; i++)
        {
            for(int j = 0; j < N; j++)
                d.insert(a[i] + b[j]);
        }
        scanf("%d", &S);
        printf("Case %d:\n", k++);
        while(S--)
        {
            scanf("%d", &X);
            flag = false;
            for(int i = 0; i < M; i++)
            {
                int ans = X - c[i];//我们要搜索的结果
                set<int>::iterator pos;
                pos = d.find(ans);
                if(pos != d.end())
                {
                    flag = true;
                    break;
                }
            }
            if(!flag)
                printf("NO\n");
            else
                printf("YES\n");
        }
    }
    return 0;
}

代码二：用set存储另外一个数组，结果TLE。。。

#include <cstdio>
#include <cstring>
#include <set>
#include <algorithm>
#define MAXN 500+1
using namespace std;
int a[MAXN], b[MAXN], c[MAXN];
int e[2500000+10];
set<int> d;
int main()
{
    int L, N, M, S, X;
    int k = 1;
    int top;
    bool flag;
    while(scanf("%d%d%d", &L, &N, &M) != EOF)
    {
        d.clear();
        for(int i = 0; i < L; i++)
            scanf("%d", &a[i]);
        for(int i = 0; i < N; i++)
            scanf("%d", &b[i]);
        for(int i = 0; i < M; i++)
            scanf("%d", &c[i]), d.insert(c[i]);
        top = 0;
        for(int i = 0; i < L; i++)
        {
            for(int j = 0; j < N; j++)
                e[top++] = a[i] + b[j];
        }
        scanf("%d", &S);
        printf("Case %d:\n", k++);
        while(S--)
        {
            scanf("%d", &X);
            flag = false;
            for(int i = 0; i < top; i++)
            {
                int ans = X - e[i];//我们要搜索的结果
                set<int>::iterator pos;
                pos = d.find(ans);
                if(pos != d.end())
                {
                    flag = true;
                    break;
                }
            }
            if(!flag)
                printf("NO\n");
            else
                printf("YES\n");
        }
    }
    return 0;
}