PAT1121:Damn Single

1121. Damn Single (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

"Damn Single (单身狗)" is the Chinese nickname for someone who is being single. You are supposed to find those who are alone in a big party, so they can be taken care of.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=50000), the total number of couples. Then N lines of the couples follow, each gives a couple of ID's which are 5-digit numbers (i.e. from 00000 to 99999). After the list of couples, there is a positive integer M (<=10000) followed by M ID's of the party guests. The numbers are separated by spaces. It is guaranteed that nobody is having bigamous marriage (重婚) or dangling with more than one companion.

Output Specification:

First print in a line the total number of lonely guests. Then in the next line, print their ID's in increasing order. The numbers must be separated by exactly 1 space, and there must be no extra space at the end of the line.

Sample Input:

3
11111 22222
33333 44444
55555 66666
7
55555 44444 10000 88888 22222 11111 23333

Sample Output:

5
10000 23333 44444 55555 88888

思路

给你n对情侣的代号以及一场派对的代号名单，找出名单中的单身狗（情侣中只要有一个没在也算），代号升序输出。

1.用一个isExist数组标记伴侣的存在，比如isExist[i] == 1 表示代号为i的人的伴侣也在派对上。

2.用couple数组关联两个人，guest数组记录在排队上的人。

3.遍历guest进行相关处理即可。

注意：1)用map关联一对情侣会超时，改用数组直接关联。

2)set的迭代器如果先被声明在循环外部（处理第一个数据的空格字符），然后再进入迭代循环会超时，所以得直接在迭代内部声明迭代器。

（这个问题不是很清楚出在哪，猜测可能跟迭代器的构造有关）

代码

#include <cstdio>
#include <set>
#include <vector>
using namespace std;
int main() {
    int n, a, b, m;
    scanf("%d", &n);
    vector<int> couple(100000);
    for (int i = 0; i < 100000; i++)
        couple[i] = -1;
    for (int i = 0; i < n; i++) {
        scanf("%d%d", &a, &b);
        couple[a] = b;
        couple[b] = a;
    }
    scanf("%d", &m);
    vector<int> guest(m), isExist(100000);
    for (int i = 0; i < m; i++) {
        scanf("%d", &guest[i]);
        if (couple[guest[i]] != -1) {
            isExist[couple[guest[i]]] = 1;
        }
    }
    set<int> s;
    for (int i = 0; i < m; i++) {
        if (!isExist[guest[i]]) {
            s.insert(guest[i]);
        }
    }
    printf("%d\n", s.size());
    for (set<int>::iterator it = s.begin(); it != s.end(); it++) {
        if (it != s.begin())
            printf(" ");
        printf("%05d", *it);
    }
    return 0;
}