UVA11419 SAM I AM

UVA11419 SAM I AM

给定一个 \(R\times C\) 的矩阵中的 \(N\) 个点，求最少选取多少个行或列才能使得每个给出的点都被一行或一列覆盖，输出方案

\(R,\ C\leq10^3,\ N\leq10^6\)

网络流

---

易知原题即为建出二分图后跑最小点覆盖求方案

这里只是记录一下如何求方案……

找出从原点开始，只经过没被使用的边，构成的连通块

如果左侧节点没被包含在连通块中，输出方案

如果右侧节点被包含在连通块中，输出方案

代码

#include <bits/stdc++.h>
using namespace std;

const int maxn = 2010, inf = INT_MAX;
int R, C, N;
bool vis[maxn], isv[maxn];
int S, T, cnt, h[maxn], cur[maxn], dis[maxn];

struct edges {
  int nxt, to, w;
  edges(int x = 0, int y = 0, int z = 0) :
    nxt(x), to(y), w(z) {}
} e[1010010];

void addline(int u, int v, int w) {
  e[++cnt] = edges(h[u], v, w), h[u] = cnt;
  e[++cnt] = edges(h[v], u, 0), h[v] = cnt;
}

bool bfs() {
  static int Q[maxn];
  memcpy(cur, h, sizeof h);
  memset(dis, 0, sizeof dis);
  int l = 1, r = 1;
  Q[1] = S, dis[S] = 1;
  while (l <= r) {
    int u = Q[l++];
    for (int i = h[u]; i; i = e[i].nxt) {
      int v = e[i].to;
      if (e[i].w && !dis[v]) {
        Q[++r] = v, dis[v] = dis[u] + 1;
        if (v == T) return 1;
      }
    }
  }
  return 0;
}

int dfs(int u, int f) {
  if (u == T || !f) {
    return f;
  }
  int res = 0, tmp;
  for (int& i = cur[u]; i && f; i = e[i].nxt) {
    int v = e[i].to;
    if (e[i].w && dis[v] == dis[u] + 1) {
      if (!(tmp = dfs(v, min(f, e[i].w)))) {
        dis[v] = 0; continue;
      }
      res += tmp, f -= tmp, e[i].w -= tmp, e[i ^ 1].w += tmp;
    }
  }
  return res;
}

int dinic() {
  int res = 0;
  while (bfs()) {
    res += dfs(S, inf);
  }
  return res;
}

void find(int u) {
  vis[u] = 1;
  for (int i = h[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (e[i].w && !vis[v]) find(v);
  }
}

void solve() {
  cnt = 1;
  memset(h, 0, sizeof h);
  memset(isv, 0, sizeof isv);
  memset(vis, 0, sizeof vis);
  S = R + C + 1, T = S + 1;
  for (int i = 1; i <= R; i++) {
    addline(S, i, 1);
  }
  for (int i = 1; i <= C; i++) {
    addline(R + i, T, 1);
  }
  for (int i = 1, x, y; i <= N; i++) {
    scanf("%d %d", &x, &y);
    isv[x] = isv[R + y] = 1;
    addline(x, R + y, 1);
  }
  printf("%d", dinic());
  find(S);
  for (int i = 1; i <= R; i++) {
    if (isv[i] && !vis[i]) printf(" r%d", i);
  }
  for (int i = R + 1; i <= R + C; i++) {
    if (isv[i] && vis[i]) printf(" c%d", i - R);
  }
  putchar(10);
}

int main() {
  while (~scanf("%d %d %d", &R, &C, &N) && R && C && N) {
    solve();
  }
  return 0;
}