ZOJ 3601 Unrequited Love 浙江省第九届省赛

Unrequited Love

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Time Limit: 16 Seconds      Memory Limit: 131072 KB

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There are n single boys and m single girls. Each of them may love none, one or several of other people unrequitedly and one-sidedly. For the coming q days, each night some of them will come together to hold a single party. In the party, if someone loves all the others, but is not loved by anyone, then he/she is called king/queen of unrequited love.

Input

There are multiple test cases. The first line of the input is an integer T ≈ 50 indicating the number of test cases.

Each test case starts with three positive integers no more than 30000 -- n m q. Then each of the next n lines describes a boy, and each of the next m lines describes a girl. Each line consists of the name, the number of unrequitedly loved people, and the list of these people's names. Each of the last q lines describes a single party. It consists of the number of people who attend this party and their names. All people have different names whose lengths are no more than 20. But there are no restrictions that all of them are heterosexuals.

Output

For each query, print the number of kings/queens of unrequited love, followed by their names in lexicographical order, separated by a space. Print an empty line after each test case. See sample for more details.

Sample Input

2
2 1 4
BoyA 1 GirlC
BoyB 1 GirlC
GirlC 1 BoyA
2 BoyA BoyB
2 BoyA GirlC
2 BoyB GirlC
3 BoyA BoyB GirlC
2 2 2
H 2 O S
He 0
O 1 H
S 1 H
3 H O S
4 H He O S

Sample Output

0
0
1 BoyB
0

0
0

看的人家的代码算是理解了，我是直接比较的明显要超时（n^2）,换个思路后时间复杂度变为（2*n）代码：

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<map>
 5 #include<set>
 6 using namespace std;
 7 char s1[30],s2[30],s3[30];
 8 int main()
 9 {
10    int T;
11    int n,m,q,n0;
12    scanf("%d",&T);
13    while(T--)
14    {
15        string lovy,lovyed;
16        map< pair<string,string> ,int >love;
17        love.clear();
18        scanf("%d %d %d",&n,&m,&q);
19        lovy = lovyed = "";
20        while(n--)
21        {
22         scanf("%s %d",s1,&n0);
23            lovy = s1;
24            while(n0--)
25            {
26                scanf("%s",s2);
27                lovyed = s2;
28                love[make_pair(lovy,lovyed)] = 1;
29            }
30            lovy = lovyed = "";
31        }
32 
33        while(m--)
34        {
35           scanf("%s %d",s1,&n0);
36            lovy = s1;
37            while(n0--)
38            {
39                scanf("%s",s2);
40                lovyed = s2;
41                love[make_pair(lovy,lovyed)] = 1;
42            }
43         lovy = lovyed = "";
44        }
45        while(q--)
46        {   bool flag = true;
47            int k =0;
48            string mm[30010];
49            scanf("%d",&n0);
50            scanf("%s",s1);
51            mm[0]=lovy = s1;
52            for(int i=1;i<n0;i++)
53            {
54              scanf("%s",s1);
55              lovyed = mm[i] = s1;
56              //关键在于下面的判断
57              if(love[make_pair(lovy,lovyed)] == 0|| love[make_pair(lovyed,lovy)] == 1)
58              {
59                     lovy = mm[i];
60                     k = i;
61              }
62            }
63            for(int j=0;j<k;j++)
64            {
65              if(lovy != mm[j])
66              {
67                if(love[make_pair(lovy,mm[j])] == 0 || love[make_pair(mm[j],lovy)] == 1)
68                   {
69                                     flag =  false;
70                   }
71              }
72            }
73            if(flag)
74            printf("1 %s\n",lovy.c_str());
75                else printf("0\n");
76        }
77      printf("\n");
78    }
79   return 0;
80 }