Warm up 2
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 243 Accepted Submission(s): 108
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.
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.
思路：我们可以将这些多米诺骨牌看成点，若两个多米诺骨牌重合，则将它们两之间连一条边，建图之后，我们可以发现，这个图只可能由下列三种情况组成，一个单独的点，一条链，一个环，对于单独的点，说明没有其他骨牌和它重合，答案加1即可，对于一条链，我们可以发现这条链一定是由水平的骨牌和竖直的骨牌依次相连形成，这时我们可以通过将同方向的一种骨牌全部拿掉使得没有重合的骨牌（因为相同方向的骨牌不可能重合），这时设链的长度为S，则答案加 （S+1）/2即可，对于环的情况与链的情况相同。；另外二分图最大匹配也可以做。
下面是参考代码：
#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#define maxn 110
using namespace std;
int map[maxn][maxn];
int vis[2010];
struct edge
{
int to;
int next;
}e[20010];
int box[2010],cnt;
void init()
{
cnt=0;
memset(box,-1,sizeof(box));
}
void add(int from,int to)
{
e[cnt].to=to;
e[cnt].next=box[from];
box[from]=cnt++;
}
int dfs(int now)
{
vis[now]=1;
int num=1;
for(int t=box[now];t+1;t=e[t].next)
{
int v=e[t].to;
if(!vis[v])
num+=dfs(v);
}
return num;
}
int main()
{
//freopen("dd.txt","r",stdin);
int n,m;
while(scanf("%d%d",&n,&m)&&(m+n))
{
int i,x,y;
init();
memset(map,0,sizeof(map));
for(i=1;i<=n;i++)
{
scanf("%d%d",&x,&y);
map[x][y]=map[x+1][y]=i;
}
for(i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
if(map[x][y])
{
add(map[x][y],i+n);
add(i+n,map[x][y]);
}
if(map[x][y+1])
{
add(map[x][y+1],i+n);
add(i+n,map[x][y+1]);
}
}
int ans=0;
memset(vis,0,sizeof(vis));
for(i=1;i<=n+m;i++)
{
if(!vis[i])
{
int tmp=dfs(i);
ans+=(tmp+1)/2;
}
}
printf("%d\n",ans);
}
return 0;
}
