Codeforces Round #277.5 (Div. 2)---B. BerSU Ball （贪心）

BerSU Ball

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls
are already busy rehearsing waltz, minuet, polonaise and quadrille moves.

We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.

For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys
and m girls.

Input

The first line contains an integer n (1 ≤ n ≤ 100)
— the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100),
where ai is
the i-th boy's dancing skill.

Similarly, the third line contains an integer m (1 ≤ m ≤ 100)
— the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100),
where bj is
the j-th girl's dancing skill.

Output

Print a single number — the required maximum possible number of pairs.

Sample test(s)

input

4
1 4 6 2
5
5 1 5 7 9

output

3

input

4
1 2 3 4
4
10 11 12 13

output

0

input

5
1 1 1 1 1
3
1 2 3

output

2

题意：有n个boy，m个girl，每一个人都有自己的舞蹈技术等级，现规定仅仅有boy和girl的等级相差不大于1才干构成一对舞伴，在每一个人都不反复的情况下，问最多能构成多少对？

解题思路：贪心。把两个数组都从小到大排序，再依次用当前最小的去跟对方比，若符合条件，则两方下标都++；若自己太低，则自己下标++，否则对方下标++。

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff

int a[105], b[105];       //boy，girl

int main()
{
    #ifdef sxk
        freopen("in.txt","r",stdin);
    #endif
    int n, m;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0; i<n; i++)
            scanf("%d", &a[i]);
        scanf("%d",&m);
        for(int i=0; i<m; i++)
            scanf("%d", &b[i]);
        sort(a, a+n);              //排序
        sort(b, b+m);
        int ans = 0;
        for(int i=0, j=0; i<n && j<m;){
            if(abs(a[i]-b[j])<=1) {     //符合条件
                ans++;
                i++;
                j++;
            }
            else if(a[i] > b[j]){        //对方太低
                j++;
            }
            else  i++;                   //自己太低
        }
        printf("%d\n", ans);
    }
    return 0;
}