# Preface

Bloggers are stupid , If there is any mistake , Welcome to point out the discussion in the comments section .

The maximum matching use of bipartite graphs \(Dinic\) Algorithm implementation , The time complexity is \(O(n\sqrt{e})\), among , \(n\) Is the number of left vertices in a bipartite graph , \(e\) Is the number of edges in a bipartite graph . If it's the Hungarian algorithm , The time complexity is \(O(nm)\) , \(m\) Is the number of right vertices in a bipartite graph , Not recommended .

Links to examples in the article .

# König Theorem

Theorem content ： The number of points covered by the smallest point of a bipartite graph ** be equal to ** The number of edges of a bipartite graph with maximum matching .

Construction method \(+\) Simple proof ：

First, find the maximum matching in bipartite graph , It is recommended to use \(Dinic\) .

Starting from every unmatched point , Traversing along the direction of the unmatched edge , Mark the points traversed backward along the matching edge . Select the left point that has not been marked , The marked dot in the right dot , Then these points can form the minimum point cover of the bipartite graph .

The traversal code is implemented as follows ：

`void dfs(int now) {`

vis[now] = true;

int SIZ = v[now].size();

for(int i = 0; i < SIZ; i++) {

int next = v[now][i].to;

if(vis[next] || !v[now][i].val)// The capacity of the positive side is 0 The description is the matching edge , The capacity of the opposite side is 0 The description is a non matching edge

continue;

dfs(next);

}

}

So it has the following properties ：

- If the point is a non matching point on the left , Then this point must be visited , Because this is the whole point \(dfs\) The starting point of the
- If the point is a non matching point on the right , This point will not be accessed , If you reach this point from the unmatched point on the left , Then you can make this edge a matching edge , Then the matching number \(+1\) , Conflict with maximum match . If the matching point on the left reaches this point , So the path of this point is
**Left unmatched point**→**Right matching point**→**Left unmatched point**→**Right matching point**→**……**→**The matching point on the left**→**Right mismatches**, Obviously , The above path is Zengguang road , Conflict with maximum match . therefore , The mismatches on the right must not be accessed . - For a set of matching points , Or both are marked , Or they're not marked . Because the matching point on the left is traversed by the matching point on the right , There must be a couple .

With the above three properties , You can find ： Select the unmarked points in the left point according to , The rule of the marked point in the right dot , The number of selected points must be the maximum number of matching edges . The unmatched point on the left must be accessed , You will not be elected , The mismatches on the right must not be accessed , You will not be elected . And the third nature determines , For a set of matching points , Will choose to have and only have one point . Therefore, the number of selected points is equal to the maximum number of matching edges .

Second, we need to solve a problem ： Make sure these points cover all the edges . It can be divided into four categories ：

- The left part is the unmatched point , The right part is the unmatched point . Property two has been discussed , It's not possible , It does not satisfy the premise of maximum matching .
- The left is the matching point , The right part is the unmatched point . The same is true for property two , The path is similar to , There's going to be an augmented road , Then the matching point on the left must not have been accessed , Must be chosen .
- The left is the matching point , The right part is the matching point . One of the matching points must be selected .
- The left part is the unmatched point , The right part is the matching point . This edge is unmatched , And the starting point is the unmatched dot on the left , So this point on the right must have been visited , Must be chosen .

Finally, make sure it's the smallest solution ： There's only one point on each side , There is no waste .

Above , Certificate completion .

Title source ：COCI 2019/2020 Contest #6 T4. Skandi

# The main idea of the topic

Given a \(n\times m\) Matrix , The white dot is \(0\) , The black dots are \(1\) . The black dots can go down to the bottom , Turn white dots into blue dots , Until the black dot . Empathy , It can also be extended to the right . How many times can the whole matrix be expanded to the point where there is no white , And print out where each expansion started , And print out the expansion direction . The first line and the first column must be black dots .

# Ideas

A modeling problem .

There are only two ways to turn a white dot into a blue dot , From the black dot above or to the left , And it only needs one point to expand . We can consider the minimum point coverage problem .

Because for a black dot , It can expand to the right or down . So it has two identities , That is to say, a point has two numbers . A number is the order in which the entire matrix is pulled into a chain , The other number is the previous number \(+n\times m\) , So there's no conflict . Get the numbered function ：

`int GetHash(int i, int j) {`

return (i - 1) * m + j;

}

So it's not hard to find a white dot , Associated with it is a number \(\leqslant n\times m\) The point of , And a number \(>n\times m\) The point of . Connect these two points , It's a bipartite graph .

The problem is to find the minimum point cover of this graph . Use \(Dinic\) , According to the above \(König\) Theorem construction can be .

The number of sides is the number of white dots , The left dot is the number of black dots , Then the time complexity is \(O(nm\sqrt{nm})\) , namely \(O(n^{\frac{3}{2}}m^{\frac{3}{2}})\) , Topic \(n\) , \(m\) All less than \(500\) , Probably in \(1s\) Find out the answer in .

# C++ Code

`#include <queue>`

#include <cstdio>

#include <vector>

#include <cstring>

#include <iostream>

using namespace std;

#define INF 0x3f3f3f3f

const int MAXN = 1e6 + 5;

const int MAXM = 5e2 + 5;

struct Node {

int to, val, rev;// In turn ： The next point , Edge capacity , The number of the opposite side

Node() {}

Node(int T, int V, int R) {

to = T;

val = V;

rev = R;

}

};

vector<Node> v[MAXN];// use vector The habit of saving pictures ...

int dn[MAXN], rt[MAXN];// Preprocessing white dots can be extended from the right two dots

queue<int> q;

int de[MAXN], be[MAXN];

int twin[MAXN];

bool vis[MAXN];

int n, m, s, t;

int arr[MAXM][MAXM];

bool bfs() {// Layer the residual network

bool flag = 0;

memset(de, 0, sizeof(de));

while(!q.empty())

q.pop();

q.push(s);

de[s] = 1; be[s] = 0;

while(!q.empty()) {

int now = q.front();

q.pop();

int SIZ = v[now].size();

for(int i = 0; i < SIZ; i++) {

int next = v[now][i].to;

if(v[now][i].val && !de[next]) {

q.push(next);

be[next] = 0;

de[next] = de[now] + 1;

if(next == t)

flag = 1;

}

}

}

return flag;

}

int dfs(int now, int flow) {// Along the Zengguang road

if(now == t || !flow)

return flow;

int i, surp = flow;

int SIZ = v[now].size();

for(i = be[now]; i < SIZ && surp; i++) {

be[now] = i;

int next = v[now][i].to;

if(v[now][i].val && de[next] == de[now] + 1) {

int maxnow = dfs(next, min(surp, v[now][i].val));

if(!maxnow)

de[next] = 0;

v[now][i].val -= maxnow;

v[next][v[now][i].rev].val += maxnow;

surp -= maxnow;

}

}

return flow - surp;

}

int Dinic() {// Network maximum flow , It can also be used for bipartite graph matching

int res = 0;

int flow = 0;

while(bfs())

while(flow = dfs(s, INF))

res += flow;

return res;

}

int GetHash(int i, int j) {// Get the number of the point

return (i - 1) * m + j;

}

void Down(int now, int i, int j) {// The black dot extends down , Each white dot can be traversed up to one time at most

if(i != now)

dn[GetHash(now, j)] = GetHash(i, j);

if(arr[now + 1][j] == 2)

Down(now + 1, i, j);

}

void Right(int now, int i, int j) { // The black dot extends to the right , Each white dot can be traversed up to one time at most

if(j != now)

rt[GetHash(i, now)] = GetHash(i, j) + n * m;

if(arr[i][now + 1] == 2)

Right(now + 1, i, j);

}

void GetMin(int now) {//dfs Find the way of construction

vis[now] = true;

int SIZ = v[now].size();

for(int i = 0; i < SIZ; i++) {

int next = v[now][i].to;

if(vis[next] || !v[now][i].val)

continue;

GetMin(next);

}

}

int main() {

scanf("%d %d", &n, &m);

s = 0; t = 2 * n * m + 1;// Source and sink initialization

char ch;

for(int i = 1; i <= n; i++) {

for(int j = 1; j <= m; j++) {

cin >> ch;

if(ch == '1')

arr[i][j] = 1;

else

arr[i][j] = 2;

}

}

for(int i = 1; i <= n; i++) {

for(int j = 1; j <= m; j++) {

if(i == 1 && j == 1)

continue;

if(arr[i][j] == 1) {// Right or down , A white spot will be visited 2 Time

Down(i, i, j);

Right(j, i, j);

}

}

}

for(int i = 1; i <= n; i++)

for(int j = 1; j <= m; j++) {

if(arr[i][j] == 1) {// Source to left , The meeting point to the right is connected to the side

int now = GetHash(i, j);

int idnow = v[now].size();

int ids = v[s].size();

v[s].push_back(Node(now, 1, idnow));

v[now].push_back(Node(s, 0, ids));

now = GetHash(i, j) + n * m;

idnow = v[now].size();

int idt = v[t].size();

v[now].push_back(Node(t, 1, idt));

v[t].push_back(Node(now, 0, idnow));

}

}

for(int i = 1; i <= n; i++) {

for(int j = 1; j <= m; j++) {

if(i == 1 && j == 1)

continue;

if(arr[i][j] == 1)

continue;

int A = dn[GetHash(i, j)];// From the left to the right

int B = rt[GetHash(i, j)];

int idA = v[A].size();

int idB = v[B].size();

v[A].push_back(Node(B, 1, idB));

v[B].push_back(Node(A, 0, idA));

}

}

printf("%d\n", Dinic());

GetMin(s);

for(int i = 1; i <= n; i++) {

for(int j = 1; j <= m; j++) {

if(arr[i][j] == 2)

continue;

if(!vis[GetHash(i, j)])// Print answers

printf("%d %d DOLJE\n", i, j);

if(vis[GetHash(i, j) + n * m])

printf("%d %d DESNO\n", i, j);

}

}

return 0;

}

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