SAM I AM UVA

就是棋盘问题输出一组解

https://blog.csdn.net/llx523113241/article/details/47759745

http://www.matrix67.com/blog/archives/116

求最小覆盖的步骤大致如下：1）在右边找到一个未被匹配过的点，标记。2）走一条没被匹配过的边，到左边的点，标记。3）走一条匹配过的边到右边，标记。4）重复2,3步骤直到不能再走。5）回到步骤一，直到找不到未被匹配且未被标记的右边的点。6）标记结束后，右边没有标记的点，和左边标记过的点，就可以覆盖所有的边。

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 1010, INF = 0x7fffffff;
int line[maxn][maxn], girl[maxn], used[maxn], boy[maxn], cx[maxn], cy[maxn];
int n, m, c;

bool find(int x)
{
    for(int i=1; i<=m; i++)
    {
        if(line[x][i] == true && used[i] == false)
        {
            used[i] = 1;
            if(girl[i] == 0 || find(girl[i]))
            {
                girl[i] = x;
                boy[x] = i;
                return true;
            }
        }
    }
    return false;
}

bool dfs(int y)
{
    cy[y] = 1;
    for(int i=1; i<=n; i++)
    {
        if(line[i][y] && !cx[i])
        {
            cx[i] = 1;
            if(boy[i] == 0 || dfs(boy[i]))
                return true;
        }
    }
    return false;
}

int main()
{
    while(scanf("%d%d%d", &n, &m, &c) && n+m+c)
    {
        int res = 0;
        mem(line, 0);
        mem(girl, 0);
        mem(boy, 0);
        mem(cx, 0);
        mem(cy, 0);
        mem(inx, 0);
        mem(iny, 0);
        rap(i, 1, c)
        {
            int a, b;
            scanf("%d%d", &a, &b);
            line[a][b] = 1;
        }
        for(int i=1; i<=n; i++)
        {
            mem(used, 0);
            if(find(i))
                res++;
        }
        for(int i=1; i<=m; i++)
            if(!girl[i]) dfs(i);
        printf("%d", res);
        rap(i, 1, n) if(cx[i]) printf(" r%d", i);
        rap(i, 1, m) if(!cy[i]) printf(" c%d", i);
        printf("
");

    }

    return 0;
}