CodeForces 489B BerSU Ball

题目：

BerSU Ball

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Status

Practice

Description
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 Input
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：
给出男生人数，每个人的舞蹈值，女生人数，每个人的舞蹈值，让你配对男女，使得组出来的对数最多。要求，男女之间舞蹈差值最多为1。

思路很简单，直接二重循环就好了

总结：
看了这题，进入了一种魔怔状态，为啥我理解成了男女最多只能有一对的舞蹈值相等呢，WA了几发。。。

代码：

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <queue>using namespace std;int n, m, ans;const int maxn = 150;int boy[maxn];int girl[maxn];bool vis[maxn];bool woc[maxn];int main(){    int n, m;    //freopen("input.txt", "r", stdin);    //freopen("output.txt", "w", stdout);        while(scanf("%d", &n)!=EOF){        memset(boy, 0, sizeof(boy));        memset(girl, 0, sizeof(girl));        memset(vis, false, sizeof(vis));        memset(woc, false, sizeof(woc));        for(int i=0; i<n; i++)            scanf("%d", &boy[i]);        scanf("%d", &m);        for(int j=0; j<m; j++)            scanf("%d", &girl[j]);        sort(boy, boy+n);        sort(girl, girl+m);        int cnt = 0;        for(int i=0; i<n; i++){            for(int j=0; j<m; j++){                if(!woc[i] && !vis[j] && (girl[j]==boy[i] || girl[j]==boy[i]+1 || girl[j]==boy[i]-1)){                    cnt++;                    woc[i] = true;                    vis[j] = true;                }            }        }        printf("%d\n", cnt);    }    return 0;}