HDU 3829 cat vs dog [maximum vertex independence set (Good Graph creation )]

Note:

N cat m dogs and K bear children

Every bear child has a favorite animal. If you like cat, you must hate dog. If you like Dag, you must hate cat.

If the animal you don't like is removed and the animal you like is not removed, the bear child will be happy.

Now let you remove some animals that make the bear child happy with the most people, output the number of people

 

Analysis:

If we find it impossible to build edges between cat and dog

Finally, the number of bear children is obtained.

It's nothing more than a bear child.

That is to say, there is a relationship between happy and unhappy bear children.

Take the bear child as a vertex

Create an edge if two bear-Child animals conflict with each other

Finally, we can find the maximum point to be independent.

 

Code:

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 using namespace std; 6  7 const int maxn = 505; 8 const int INF = 1000000000; 9 10 int vis[maxn];11 int Link[maxn];12 int n;13 vector<int> G[maxn];14 int Find(int u) {15     for(int i = 0; i < G[u].size(); i++) {16         int v = G[u][i];17         if(!vis[v]) {18             vis[v] = 1;19             if(Link[v] == -1 || Find(Link[v]) ) {20                 Link[v] = u;21                 return true;22             }23         }24     }25     return false;26 }27 28 int solve() {29     memset(Link, -1, sizeof(Link));30     int cnt = 0;31     for(int i = 1; i <= n; i++) {32         if(G[i].size() ) {33             memset(vis, 0, sizeof(vis));34             if(Find(i) ) cnt++;35         }36     }37     return cnt;38 }39 40 char s1[maxn][5], s2[maxn][5];41 bool check(int i, int j) {42     if(strcmp(s1[i], s2[j]) == 0 || strcmp(s2[i], s1[j]) == 0 ) {43         return true;44     }45     return false;46 }47 48 int main() {49     int x, y;50     while(EOF != scanf("%d %d %d",&x, &y, &n) ) {51         for(int i = 1; i <= n; i++) {52             scanf("%s %s", s1[i], s2[i]);53         }54         for(int i = 1; i <= n; i++) {55             G[i].clear();56             for(int j = 1; j <= n; j++) {57                 if(i == j) continue;58                 if(check(i, j) ) {59                     G[i].push_back(j);60                 }61             }62         }63         printf("%d\n",(2 * n - solve()) / 2);64     }65     return 0;66 }

View code

 

HDU 3829 cat vs dog [maximum vertex independence set (Good Graph creation )]