hdu2768-Cat vs. Dog：图论：二分匹配

Cat vs. Dog

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1602    Accepted Submission(s): 606

Problem Description

The latest reality show has hit the TV: ``Cat vs. Dog''. In this show, a bunch of cats and dogs compete for the very prestigious Best Pet Ever title. In each episode, the cats and dogs get to show themselves off, after which the viewers vote on which pets should stay and which should be forced to leave the show.

Each viewer gets to cast a vote on two things: one pet which should be kept on the show, and one pet which should be thrown out. Also, based on the universal fact that everyone is either a cat lover (i.e. a dog hater) or a dog lover (i.e. a cat hater), it has been decided that each vote must name exactly one cat and exactly one dog.

Ingenious as they are, the producers have decided to use an advancement procedure which guarantees that as many viewers as possible will continue watching the show: the pets that get to stay will be chosen so as to maximize the number of viewers who get both their opinions satisfied. Write a program to calculate this maximum number of viewers.

 

 

Input

On the first line one positive number: the number of testcases, at most 100. After that per testcase:

* One line with three integers c, d, v (1 ≤ c, d ≤ 100 and 0 ≤ v ≤ 500): the number of cats, dogs, and voters.
* v lines with two pet identifiers each. The first is the pet that this voter wants to keep, the second is the pet that this voter wants to throw out. A pet identifier starts with one of the characters `C' or `D', indicating whether the pet is a cat or dog, respectively. The remaining part of the identifier is an integer giving the number of the pet (between 1 and c for cats, and between 1 and d for dogs). So for instance, ``D42'' indicates dog number 42.

 

 

Output

Per testcase:

* One line with the maximum possible number of satisfied voters for the show.

 

 

Sample Input

2 1 1 2 C1 D1 D1 C1 1 2 4 C1 D1 C1 D1 C1 D2 D2 C1

 

 

Sample Output

1 3

 

 

Source

NWERC 2008

---

算法：将喜欢猫和喜欢狗的人分开，就形成了一个二分图，这样喜欢猫（狗）的人一定不会冲突的，然后扫描一次，将互相矛盾的人连一条边。

二分图的最大独立集就是答案。由已知定理有：二分图最大独立集=顶点数-二分图最大匹配。求二分图最大匹配即可。

 

 

 1 const int MAXN = 510;

 2 int uN,vN;//u,v的数目，使用前面必须赋值

 3 int g[MAXN][MAXN];//邻接矩阵

 4 int linker[MAXN];

 5 bool used[MAXN];

 6 #include <memory.h>

 7 bool dfs(int u)

 8 {

 9     for(int v = 0; v < vN; v++)

10         if(g[u][v] && !used[v])

11         {

12             used[v] = true;

13             if(linker[v] == -1 || dfs(linker[v]))

14             {

15                 linker[v] = u;

16                 return true;

17             }

18         }

19     return false;

20 }

21 int hungary()

22 {

23     int res = 0;

24     memset(linker,-1,sizeof(linker));

25     for(int u = 0; u < uN; u++)

26     {

27         memset(used,false,sizeof(used));

28         if(dfs(u))res++;

29     }

30     return res;

31 }

32 

33 #include <iostream>

34 #include <stdio.h>

35 #include <string.h>

36 using namespace std;

37 

38 

39 int getint(char s[])

40 {

41     int ans=0;

42     for(int i=0; i<strlen(s); i++)

43     {

44         ans=ans*10+s[i]-'0';

45     }

46     return ans;

47 }

48 

49 typedef pair<int,int> CPair;

50 

51 CPair upair[510],vpair[510];

52 int main()

53 {

54     #ifndef ONLINE_JUDGE

55     freopen("in.txt","r",stdin);

56     #endif // ONLINE_JUDGE

57     int T;

58     cin>>T;

59     while(T--)

60     {

61         memset(g,0,sizeof g);

62         int nc,nd,n;

63         uN=vN=0;

64         scanf("%d%d%d",&nc,&nd,&n);

65         for(int i=0; i<n; i++) // 人的数量

66         {

67             char s1[20],s2[20];

68             scanf("%s%s",s1,s2); // C1 D1

69             int ci=getint(s1+1),di=getint(s2+1);

70             if(s1[0]=='C')

71             {

72                 // 放到第一个集合

73                 upair[uN++]=CPair(ci,di);

74             }

75             else

76             {

77                 vpair[vN++]=CPair(ci,di);

78             }

79         }

80         // 构图

81         for(int i=0;i<uN;i++)

82             for(int j=0;j<vN;j++)

83             {

84                 // 矛盾的话就连一条边

85                 if(upair[i].first==vpair[j].second || upair[i].second==vpair[j].first)

86                     g[i][j]=1;

87             }

88 

89         int ans=uN+vN-hungary();

90         printf("%d\n",ans);

91     }

92     return 0;

93 }