UVa 11419 SAM I AM (最小覆盖数)

题意：给定一个 n * m 的矩阵，有一些格子有目标，每次可以消灭一行或者一列，问你最少要几次才能完成。

析：把 行看成 X，把列看成是 Y，每个目标都连一条线，那么就是一个二分图的最小覆盖数，这个答案就是二分图的最大匹配，在输出解的时候，就是从匈牙利树上，从X的未盖点出发，然后标记X和Y，最后X中未标记的和Y标记的就是答案。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
 
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2000 + 10;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
 
struct Edge{
  int to, next;
};
Edge edge[maxn*maxn/4];
int head[maxn], cnt;
int match[maxn];
bool used[maxn];
bool visx[maxn], visy[maxn];
 
void addEdge(int u, int v){
  edge[cnt].to = v;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}
 
bool dfs(int u){
  used[u] = true;
  for(int i = head[u]; ~i; i = edge[i].next){
    int v = edge[i].to, w = match[v];
    if(w < 0 || !used[w] && dfs(w)){
      match[u] = v;
      match[v] = u;
      return true;
    }
  }
  return false;
}
 
void dfs1(int u){
  visx[u] = 1;
  for(int i = head[u]; ~i; i = edge[i].next){
    int v = edge[i].to;
    if(visy[v])  continue;
    visy[v] = 1;
    dfs1(match[v]);
  }
}
 
int main(){
  int r, c;
  while(scanf("%d %d %d", &r, &c, &m) == 3 && r+c+m){
    ms(head, -1);  cnt = 0;
    for(int i = 0; i < m; ++i){
      int u, v;
      scanf("%d %d", &u, &v);
      --u, --v;
      addEdge(u, v + r);
    }
    n = r + c;
    ms(match, -1);
    int ans = 0;
    for(int i = 0; i < n; ++i)  if(match[i] < 0){
      ms(used, 0);  if(dfs(i))  ++ans;
    }
    printf("%d", ans);
    ms(visx, 0);  ms(visy, 0);
    ms(used, 0);
    for(int i = 0; i < r; ++i)  if(match[i] != -1)  used[i] = 1;
    for(int i = 0; i < r; ++i)  if(!used[i])  dfs1(i);
    for(int i = 0; i < r; ++i)  if(!visx[i])  printf(" r%d", i + 1);
    for(int i = r; i < n; ++i)  if(visy[i])  printf(" c%d", i + 1 - r);
    printf("\n");
  }
  return 0;
}