★hdu 3829 Cat VS Dog (最大独立集)
2017-07-11 17:14
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Cat VS Dog
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 3765 Accepted Submission(s): 1349
Problem Description
The zoo have N cats and M dogs, today there are P children visiting the zoo, each child has a like-animal and a dislike-animal, if the child's like-animal is a cat, then his/hers dislike-animal must be a dog, and vice versa.
Now the zoo administrator is removing some animals, if one child's like-animal is not removed and his/hers dislike-animal is removed, he/she will be happy. So the administrator wants to know which animals he should remove to make maximum number of happy children.
Input
The input file contains multiple test cases, for each case, the first line contains three integers N <= 100, M <= 100 and P <= 500.
Next P lines, each line contains a child's like-animal and dislike-animal, C for cat and D for dog. (See sample for details)
Output
For each case, output a single integer: the maximum number of happy children.
Sample Input
1 1 2
C1 D1
D1 C1
1 2 4
C1 D1
C1 D1
C1 D2
D2 C1
Sample Output
1
3
HintCase 2: Remove D1 and D2, that makes child 1, 2, 3 happy.
Source
2011 Multi-University Training Contest 1 - Host by HNU
思路:
倒着想,如果都是开心的孩子们, 肯定没有A喜欢的是B讨厌的或者B喜欢的是A讨厌的,那我们就给这样矛盾的孩子之间建边,然后跑一边最大独立集, 求出最多的点,这里面两两没有边, 也就是没有矛盾的, 这些都是开心的,这样动物园就剩这些孩子喜欢的就行~
二分图真奇妙。。。
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;
const int maxn = 505;
char like[maxn][maxn], dislike[maxn][maxn];
int match[maxn], book[maxn], n, m, p;
vector<int> v[maxn];
int Find(int x)
{
for(int i = 0; i < v[x].size(); i++)
{
int to = v[x][i];
if(book[to]) continue;
book[to] = 1;
if(!match[to] || Find(match[to]))
{
match[to] = x;
return 1;
}
}
return 0;
}
int main()
{
while(~scanf("%d%d%d", &n, &m, &p))
{
memset(match, 0, sizeof(match));
for(int i = 0; i < maxn; i++)
v[i].clear();
int ans = 0;
for(int i = 1; i <= p; i++)
{
scanf(" %s %s", like[i], dislike[i]);
}
for(int i = 1; i <= p; i++)
{
for(int j = 1; j <= p; j++)
{
if(!strcmp(like[i], dislike[j]) || !strcmp(dislike[i], like[j]))
v[i].push_back(j);
}
}
for(int i = 1; i <= p; i++)
{
memset(book, 0, sizeof(book));
ans += Find(i);
}
printf("%d\n", p-ans/2);
}
return 0;
}
Cat VS Dog
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 3765 Accepted Submission(s): 1349
Problem Description
The zoo have N cats and M dogs, today there are P children visiting the zoo, each child has a like-animal and a dislike-animal, if the child's like-animal is a cat, then his/hers dislike-animal must be a dog, and vice versa.
Now the zoo administrator is removing some animals, if one child's like-animal is not removed and his/hers dislike-animal is removed, he/she will be happy. So the administrator wants to know which animals he should remove to make maximum number of happy children.
Input
The input file contains multiple test cases, for each case, the first line contains three integers N <= 100, M <= 100 and P <= 500.
Next P lines, each line contains a child's like-animal and dislike-animal, C for cat and D for dog. (See sample for details)
Output
For each case, output a single integer: the maximum number of happy children.
Sample Input
1 1 2
C1 D1
D1 C1
1 2 4
C1 D1
C1 D1
C1 D2
D2 C1
Sample Output
1
3
HintCase 2: Remove D1 and D2, that makes child 1, 2, 3 happy.
Source
2011 Multi-University Training Contest 1 - Host by HNU
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