Cat VS Dog 求二分图最大独立集
2014-07-31 16:47
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Cat VS Dog
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 125536/65536K (Java/Other)
Total Submission(s) : 22 Accepted Submission(s) : 7
Font: Times New Roman | Verdana | Georgia
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.
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)
把每个学生看做一个点,设一个同学为点i,若删掉同学i不喜欢的动物,会引起另一个同学j不快乐。用map[i][j]=map[j][i]=1把同学i和j联系起来。则求出他的最大独立集就可以求出
最多个同学快乐的数量。
#include#include#include#include#includeusing namespace std;const int N = 501;int map
; int vis
;int match
;int n, m;int P;string A
, B
;void init(){ int i, j; for (i = 0; i < N; i++){ vis[i] = 0; match[i] = -1; for (j = 0; j < N; j++){ map[i][j]
= 0; } }}bool find(int x){ int i; for (i = 1; i <= P; i++){ if (!vis[i] && map[x][i]){ vis[i] = 1; if (match[i] == -1 || find(match[i])) { match[i] = x; return true; } } } return false;}int Match(){ int ans = 0; int i; for (i = 1; i <= P; i++){ memset(vis,
0, sizeof(vis)); if (find(i)) ans++; } return ans;}int main(){ freopen("in.txt", "r", stdin); while (~scanf("%d%d%d", &n, &m, &P)){ int i,j; init(); for (i = 1; i <= P; i++){ cin >> A[i] >> B[i]; }/*for(i=1;i<=P;i++){cout<<A[i]<<" "<<B[i]<<endl;}*/ getchar();
for (i = 1; i <= P; i++){ for (j = 1; j <= P; j++){ if (B[i] == A[j]){ map[j][i]=map[i][j] = 1; } } } int ans = Match(); printf("%d\n", P-ans/2); } return 0;}
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 125536/65536K (Java/Other)
Total Submission(s) : 22 Accepted Submission(s) : 7
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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
把每个学生看做一个点,设一个同学为点i,若删掉同学i不喜欢的动物,会引起另一个同学j不快乐。用map[i][j]=map[j][i]=1把同学i和j联系起来。则求出他的最大独立集就可以求出
最多个同学快乐的数量。
#include#include#include#include#includeusing namespace std;const int N = 501;int map
; int vis
;int match
;int n, m;int P;string A
, B
;void init(){ int i, j; for (i = 0; i < N; i++){ vis[i] = 0; match[i] = -1; for (j = 0; j < N; j++){ map[i][j]
= 0; } }}bool find(int x){ int i; for (i = 1; i <= P; i++){ if (!vis[i] && map[x][i]){ vis[i] = 1; if (match[i] == -1 || find(match[i])) { match[i] = x; return true; } } } return false;}int Match(){ int ans = 0; int i; for (i = 1; i <= P; i++){ memset(vis,
0, sizeof(vis)); if (find(i)) ans++; } return ans;}int main(){ freopen("in.txt", "r", stdin); while (~scanf("%d%d%d", &n, &m, &P)){ int i,j; init(); for (i = 1; i <= P; i++){ cin >> A[i] >> B[i]; }/*for(i=1;i<=P;i++){cout<<A[i]<<" "<<B[i]<<endl;}*/ getchar();
for (i = 1; i <= P; i++){ for (j = 1; j <= P; j++){ if (B[i] == A[j]){ map[j][i]=map[i][j] = 1; } } } int ans = Match(); printf("%d\n", P-ans/2); } return 0;}
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