【HDU4619】【二分匹配】【最大匹配】

Warm up 2

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2031    Accepted Submission(s): 926

Problem Description

　　Some 1×2 dominoes are placed on a plane. Each dominoe is placed either horizontally or vertically. It's guaranteed the dominoes in the same direction are not overlapped, but horizontal and vertical dominoes may overlap with each other. You task is to remove some dominoes, so that the remaining dominoes do not overlap with each other. Now, tell me the maximum number of dominoes left on the board.

 

Input

　　There are multiple input cases.
　　The first line of each case are 2 integers: n(1 <= n <= 1000), m(1 <= m <= 1000), indicating the number of horizontal and vertical dominoes.
Then n lines follow, each line contains 2 integers x (0 <= x <= 100) and y (0 <= y <= 100), indicating the position of a horizontal dominoe. The dominoe occupies the grids of (x, y) and (x + 1, y).
　　Then m lines follow, each line contains 2 integers x (0 <= x <= 100) and y (0 <= y <= 100), indicating the position of a horizontal dominoe. The dominoe occupies the grids of (x, y) and (x, y + 1).
　　Input ends with n = 0 and m = 0.

 

Output

　　For each test case, output the maximum number of remaining dominoes in a line.

 

Sample Input

2 30 00 30 11 11 34 50 10 23 12 20 01 02 04 13 20 0

 

Sample Output

46

 

Author

SYSU

#include <iostream>#include <cstring>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <cstdlib>#include <cstdio>#include <algorithm>using namespace std;   int n,m;const int N = 1100;pair<int,int> p1[N],p2[N];vector<int> G[N];int match[N];int used[N];int uN;bool dfs(int u)  {      for(int i=0;i<G[u].size();i++)      {          if(!used[G[u][i]])          {              used[G[u][i]]=true;              if(match[G[u][i]]==-1||dfs(match[G[u][i]]))              {                  match[G[u][i]]=u;                  return true;              }          }      }      return false;  }  int hungary()  {      int u;      int res=0;      memset(match,-1,sizeof(match));      for(u=0;u<uN;u++)      {          memset(used,false,sizeof(used));          if(dfs(u)) res++;      }      return res;  }  int main(){    while(scanf("%d%d",&n,&m) != EOF){if(n == 0 && m == 0) break;for(int i=0;i<n;i++){G[i].clear();}memset(match,-1,sizeof(match));for(int i=0;i<n;i++){int x,y;scanf("%d%d",&x,&y);p1[i] = make_pair(x,y);}for(int j=0;j<m;j++){int x,y;scanf("%d%d",&x,&y);p2[j] = make_pair(x,y);}for(int i=0;i<n;i++){for(int j=0;j<m;j++){int x1 = p1[i].first;int y1 = p1[i].second;int x2 = p2[j].first;int y2 = p2[j].second;if((x1+1==x2 && y1==y2) || (x1+1==x2 && y1==y2+1) || (x1==x2 && y1==y2)|| (x1==x2 && y1==y2+1)){G[i].push_back(j);}}}//cout << "ok in 95 " << endl;uN = n;int ret = hungary();//cout << ret << endl;printf("%d\n",n+m-ret);}    return 0;}