hdu4619 - Warm up 2(联通图or贪心0r二分图匹配)【待完善】

这道题好多种做法，，，

1、联通图。

    由于题目特点，这道题可以转化为求图的该图的各个子联通图，每个包含x个点的联通图都可以最多得到不重叠的x/2个牌。

注意这句话：很重要的（It's guaranteed the dominoes in the same direction are not overlapped, but horizontal and vertical dominoes may overlap with each other.）

代码如下：

#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
#define M 2005
int n, m, u[M][2], v[M][2];
bool vis[150][150];
vector<int>g[150][150];
void dfs(int x, int y, int &tt)//遍历联通图得到节点数目
{
    vis[x][y] = 1;
    tt++;
    for(int i = 0; i < (int)g[x][y].size(); i++)
    {
        int d = g[x][y][i];
        int xx = u[d][0], yy = u[d][1];
        if(vis[xx][yy]==0)
            dfs(xx,yy,tt);
        xx = v[d][0], yy = v[d][1];
        if(vis[xx][yy]==0)
            dfs(xx,yy,tt);
    }
}
int main ()
{
    int x, y;
    while(scanf("%d %d",&n,&m), n+m)
    {
        for(int i = 0; i < 150; i++)
            for(int j = 0; j < 150; j++)
                g[i][j].clear();
        for(int i = 0; i < n; i++)
        {
            scanf("%d %d",&x, &y);
            u[i][0] = x;
            u[i][1] = y;
            v[i][0] = x+1;
            v[i][1] = y;
            g[x][y].push_back(i);
            g[x+1][y].push_back(i);
        }
        for(int i = n; i < n+m; i++)
        {
            scanf("%d %d",&x, &y);
            u[i][0] = x;
            u[i][1] = y;
            v[i][0] = x;
            v[i][1] = y+1;
            g[x][y].push_back(i);
            g[x][y+1].push_back(i);
        }
        memset(vis,0,sizeof(vis));
        int ans = 0, tt = 0;
        for(int i = 0; i < 150; i++)
            for(int j = 0; j < 150; j++)
                if(g[i][j].size()!=0&&vis[i][j]==0)
                    {tt = 0; dfs(i,j,tt); ans+=(tt/2);}
        printf("%d\n",ans);
    }
    return 0;
}