pat甲级 1121. Damn Single (25)

来源：互联网发布：户口本查询软件编辑：程序博客网时间：2024/04/27 22:44

"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:

311111 2222233333 4444455555 66666755555 44444 10000 88888 22222 11111 23333

Sample Output:

510000 23333 44444 55555 88888题目大意给出n对夫妻，再给出m个参加party的人，问这m个人中单身的人有几个并且按从小到大的顺序输出这几个人。解题思路用一个数组mp[]来存储n对夫妻的信息，用vis[]来存储参加party的人。对每个参加party的人：（1）没有配偶；（2）有配偶但是配偶没在party中；（3）有配偶并且配偶也在party中。显然对于（1）（2）是应该输出的。即mp==-1和!vis[]的人。代码#include <cstdio>#include <vector>#include <map>#include <set>#include <cstdlib>#include <climits>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define ll long longconst int MAXN = 100000 + 5;const int MAXM = 10000 + 5;const int INF = 0x7f7f7f7f;template <class XSD> inline XSD f_min(XSD a, XSD b) { if (a > b) a = b; return a; }template <class XSD> inline XSD f_max(XSD a, XSD b) { if (a < b) a = b; return a; }int n, m;int mp[MAXN], digit[MAXM];int vis[MAXN];void Getdata(int n){    int x, y;    memset(vis, 0, sizeof vis);    memset(mp, -1, sizeof mp);    for(int i=0; i<n; i++){        scanf("%d%d", &x, &y);        mp[x] = y;        mp[y] = x;    }    scanf("%d", &m);    for(int i=0; i<m; i++) {        scanf("%d", &digit[i]);        vis[digit[i]] = 1;    }}void Solve(){    int ans[MAXM], cnt=0;    for(int i=0; i<m; i++){        int x=digit[i], y=mp[x];        if(y==-1) ans[cnt++]=x;//没有伴侣        else{            if(!vis[y]) ans[cnt++]=x;//有伴侣，但是没来        }    }    printf("%d\n", cnt);    sort(ans, ans+cnt);    for(int i=0; i<cnt; i++) printf("%05d%c", ans[i], i==(cnt-1)?'\n':' ');}int main(){    while(~scanf("%d", &n)){        Getdata(n);        Solve();    }    return 0;}

1 0