PAT (Advanced Level) Practice A1121 Damn Single (25 分) (C++)(甲级)(两个数组就可以解决,时间复杂度O(N))

版权声明:假装有个原创声明……虽然少许博文不属于完全原创,但也是自己辛辛苦苦总结的,转载请注明出处,感谢! https://blog.csdn.net/m0_37454852/article/details/88529170

原题链接

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
using namespace std;

const int MAX = 100010, INF = 1<<30;
int N, M, A, B, X;
int couple[MAX];
bool isCome[MAX] = {0};
queue<int> Q;

int main()
{
    memset(couple, -1, sizeof(couple));//先认为都没有对象
    scanf("%d", &N);
    for(int i=0; i<N; i++)
    {
        scanf("%d %d", &A, &B);
        couple[A] = B;//A的对象是B,B的对象是A
        couple[B] = A;
    }
    scanf("%d", &M);
    for(int i=0; i<M; i++)
    {
        scanf("%d", &X);
        isCome[X] = 1;//X来到了party
    }
    for(int i=0; i<MAX; i++)
    {
        if(isCome[i] && (couple[i] == -1 || !isCome[couple[i]]))
        {//如果i来到了party,但他没有对象,或者他对象没有来,那他就是lonely guests
            Q.push(i);//存入队列
        }
    }
    printf("%d\n", Q.size());
    if(!Q.empty()) printf("%05d", Q.front()), Q.pop();//注意格式
    while(!Q.empty()) printf(" %05d", Q.front()), Q.pop();
	return 0;
}



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转载自blog.csdn.net/m0_37454852/article/details/88529170