HDU 4619 (二分图匹配 最大独立集)

Warm up 2

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2052    Accepted Submission(s): 938

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 3
0 0
0 3
0 1
1 1
1 3
4 5
0 1
0 2
3 1
2 2
0 0
1 0
2 0
4 1
3 2
0 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    4
6
   
   
   
   

 

题意是给你ｎ个横向的骨牌和ｍ个纵向的骨牌，保证各自方向的骨牌不相交，问最少拿掉多少张骨牌使得所有的骨牌没有相交的情况．骨牌的长度是１．

用到了二分图的一个性质，二分图的最大独立集＝所有的节点数－二分图最大匹配．

正好横向纵向两类点，如果相交就连边，然后就求个最大匹配就好了．

#include <bits/stdc++.h>
using namespace std;
#define maxn 2111
#define maxm 2111111

int n, m;
int a[maxn][2], b[maxn][2];
struct node {
    int from, to, next;
}edge[maxm];
int head[maxn];
int cnt;

void add_edge (int from , int to) {
    edge[cnt].from = from, edge[cnt].to = to, edge[cnt].next = head[from], head[from] = cnt++;
}

bool jiao (int i, int j) {
    if (a[i][0] == b[j][0] && a[i][1] == b[j][1])
        return 1;
    if (a[i][0]+1 == b[j][0] && a[i][1] == b[j][1])
        return 1;
    if (a[i][0] == b[j][0] && a[i][1] == b[j][1]+1)
        return 1;
    if (a[i][0]+1 == b[j][0] && a[i][1] == b[j][1]+1)
        return 1;
    return 0;
}


int pre[maxn], vis[maxn];
bool dfs (int u) {
    for (int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].to;
        if (!vis[v]) {
            vis[v] = 1;
            if (pre[v] == -1 || dfs (pre[v])) {
                pre[v] = u;
                return 1;
            }
        }
    }
    return 0;
}

int hungry () {
    int ans = 0;
    memset (pre, -1, sizeof pre);
    for (int i = 1; i <= n; i++) {
        memset (vis, 0, sizeof vis);
        if (dfs (i))
            ans++;
    }
    return ans;
}

int main () {
    //freopen ("in", "r", stdin);
    ios::sync_with_stdio (0);
    while (cin >> n >> m && n+m) {
        cnt = 0;
        memset (head, -1, sizeof head);
        for (int i = 1; i <= n; i++) {
            cin >> a[i][0] >> a[i][1];
        }
        for (int j = 1+n; j <= n+m; j++) {
            cin >> b[j][0] >> b[j][1];
            for (int i = 1; i <= n; i++) {
                if (jiao (i, j)) {
                    add_edge (i, j);
                    add_edge (j, i);
                }
            }
        }
        cout << n+m-hungry () << endl;
    }
    return 0;
}