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;

}