Cat vs. Dog
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 1422 Accepted
Submission(s): 534
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
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好好地一题最大独立集被我毁了...
开始思路匹配错了.后来才想清楚，人应该和人匹配，构图原则是一个人喜欢的和另一个人不喜欢的一样则两者匹配，最后求最大独立集。
最大独立集=原点数-最大匹配/2(构图时构双向图)；
1 //93MS 1780K 1160 B C++
2 #include<iostream>
3 #include<vector>
4 #include<string.h>
5 #define N 505
6 using namespace std;
7 int match[N];
8 int vis[N];
9 int n,m,k;
10 vector<int>V[N];
11 int dfs(int u)
12 {
13 for(int i=0;i<V[u].size();i++){
14 int v=V[u][i];
15 if(!vis[v]){
16 vis[v]=1;
17 if(match[v]==-1 || dfs(match[v])){
18 match[v]=u;
19 return 1;
20 }
21 }
22 }
23 return 0;
24 }
25 int hungary()
26 {
27 int ret=0;
28 memset(match,-1,sizeof(match));
29 for(int i=0;i<k;i++){
30 memset(vis,0,sizeof(vis));
31 ret+=dfs(i);
32 }
33 return ret;
34 }
35 int main(void)
36 {
37 char a[N][5],b[N][5];
38 int t;
39 scanf("%d",&t);
40 while(t--)
41 {
42 scanf("%d%d%d",&n,&m,&k);
43 for(int i=0;i<=k;i++) V[i].clear();
44 for(int i=0;i<k;i++)
45 scanf("%s%s",&a[i],&b[i]);
46 for(int i=0;i<k;i++)
47 for(int j=0;j<k;j++){
48 if(strcmp(a[i],b[j])==0 || strcmp(a[j],b[i])==0){
49 V[i].push_back(j);
50 V[j].push_back(i);
51 }
52 }
53 printf("%d\n",k-hungary()/2);
54 }
55 return 0;
56 }
hdu 2768 Cat vs. Dog (二分匹配)