把一整行或一整列看成一个点跑最大流。
code
#include <bits/stdc++.h>
using namespace std;
const int N = 80010, M = 220, inf = 0x3f3f3f3f;
int n, m, S, T;
int a[N], d[M][M], g1[M][M], g2[M][M], tot;
int h[N], e[N * 4], c[N * 4], ne[N * 4], idx;
int dep[N], cur[N];
void add(int u, int v, int w) {
e[idx] = v, c[idx] = w, ne[idx] = h[u], h[u] = idx ++ ;
e[idx] = u, c[idx] = 0, ne[idx] = h[v], h[v] = idx ++ ;
}
bool bfs() {
memset(dep, -1, sizeof dep);
queue<int> q;
q.push(S);
dep[S] = 0, cur[S] = h[S];
while (q.size()) {
int t = q.front();
q.pop();
for (int i = h[t]; ~i; i = ne[i]) {
int j = e[i];
if (dep[j] == -1 && c[i] > 0) {
dep[j] = dep[t] + 1;
cur[j] = h[j];
if (j == T) return true;
q.push(j);
}
}
}
return false;
}
int find(int u, int limit) {
if (u == T) return limit;
int flow = 0;
for (int i = cur[u]; ~i && flow < limit; i = ne[i]) {
cur[u] = i;
int j = e[i];
if (dep[j] == dep[u] + 1 && c[i] > 0) {
int t = find(j, min(c[i], limit - flow));
if (!t) dep[j] = -1;
flow += t, c[i] -= t, c[i ^ 1] += t;
}
}
return flow;
}
int dinic() {
int res = 0, flow = 0;
while (bfs()) while (flow = find(S, inf)) res += flow;
return res;
}
int main() {
S = 0, T = 80001;
scanf("%d%d", &n, &m);
memset(h, -1, sizeof h);
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
scanf("%d", &d[i][j]);
for (int i = 1; i <= n; i ++ ) {
int l = 1;
for (int j = 1; j <= m; j ++ ) {
if (d[i][j] == 2) {
if (l <= j - 1) {
++ tot;
a[tot] = i;
add(S, tot, 1);
for (int k = l; k <= j - 1; k ++ )
g1[i][k] = tot;
}
l = j + 1;
}
}
if (l <= m) {
++ tot;
a[tot] = i;
add(S, tot, 1);
for (int k = l; k <= m; k ++ )
g1[i][k] = tot;
}
}
for (int i = 1; i <= m; i ++ ) {
int l = 1;
for (int j = 1; j <= n; j ++ ) {
if (d[j][i] == 2) {
if (l <= j - 1) {
++ tot;
a[tot] = i;
add(tot, T, 1);
for (int k = l; k <= j - 1; k ++ )
g2[k][i] = tot;
}
l = j + 1;
}
}
if (l <= n) {
++ tot;
a[tot] = i;
add(tot, T, 1);
for (int k = l; k <= n; k ++ )
g2[k][i] = tot;
}
}
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
if (d[i][j] == 0)
add(g1[i][j], g2[i][j], 1);
printf("%d\n", dinic());
for (int i = 0; i < idx; i += 2) {
if (e[i] != T && e[i ^ 1] != S && c[i] == 0)
printf("%d %d\n", a[e[i ^ 1]], a[e[i]]);
}
return 0;
}