Commit d71ec36d authored by 24OI-bot's avatar 24OI-bot
Browse files

style: format markdown files with remark-lint

parent 173499b7

docs/graph/flow/max-flow.md

+113 −137
Original line number Diff line number Diff line
@@ -33,43 +33,37 @@ 
EK 算法的时间复杂度为 $O(n^2m)$（其中 $n$ 为点数，$m$ 为边数）。效率还有很大提升空间。



```cpp

#include <algorithm>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <queue>

#define INF 0x3f3f3f3f

using namespace std;

struct edge

{

struct edge {

  int v, w, next;

} e[200005];

struct node

{

struct node {

  int v, e;

} p[10005];

int head[10005], vis[10005];

int n, m, s, t, cnt = 1;

void addedge(int u,int v,int w)

{

void addedge(int u, int v, int w) {

  e[++cnt].v = v;

  e[cnt].w = w;

  e[cnt].next = head[u];

  head[u] = cnt;

}

bool bfs()

{

bool bfs() {

  queue<int> q;

  memset(p, 0, sizeof(p));

  memset(vis, 0, sizeof(vis));

  vis[s] = 1;

  q.push(s);

 while(!q.empty())

 {

  while (!q.empty()) {

    int cur = q.front();

    q.pop();

    for (int i = head[cur]; i; i = e[i].next)

   if((!vis[e[i].v])&&e[i].w)

   {

      if ((!vis[e[i].v]) && e[i].w) {

        p[e[i].v].v = cur;

        p[e[i].v].e = i;

        if (e[i].v == t) return 1;
@@ -79,24 +73,19 @@ bool bfs() 
  }

  return 0;

}

int main()

{

int main() {

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

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

 {

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

    int u, v, w;

    scanf("%d%d%d", &u, &v, &w);

    addedge(u, v, w);

    addedge(v, u, 0);

  }

  int ans = 0;

 while(bfs())

 {

  while (bfs()) {

    int minw = INF;

  for(int i=t;i!=s;i=p[i].v)

   minw=min(minw,e[p[i].e].w);

  for(int i=t;i!=s;i=p[i].v)

  {

    for (int i = t; i != s; i = p[i].v) minw = min(minw, e[p[i].e].w);

    for (int i = t; i != s; i = p[i].v) {

      e[p[i].e].w -= minw;

      e[p[i].e ^ 1].w += minw;

    }
@@ -130,27 +119,24 @@ Dinic 算法有两个优化： 
特别地，在求解二分图最大匹配问题时，可以证明 Dinic 算法的时间复杂度是 $O(n \sqrt{m})$。



```cpp

#include <algorithm>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <queue>

#define INF 0x3f3f3f3f

using namespace std;

struct edge

{

struct edge {

  int v, w, next;

} e[200005];

int n, m, s, t, cnt = 1;

int head[100005], dep[100005], vis[100005], cur[100005];

void addedge(int u,int v,int w)

{

void addedge(int u, int v, int w) {

  e[++cnt].v = v;

  e[cnt].w = w;

  e[cnt].next = head[u];

  head[u] = cnt;

}

bool bfs()

{

bool bfs() {

  queue<int> q;

  memset(dep, INF, sizeof(dep));

  memset(vis, 0, sizeof(vis));
@@ -158,17 +144,14 @@ bool bfs() 
  dep[s] = 0;

  vis[s] = 1;

  q.push(s);

 while(!q.empty())

 {

  while (!q.empty()) {

    int p = q.front();

    q.pop();

    vis[p] = 0;

    for (int i = head[p]; i; i = e[i].next)

   if(dep[e[i].v]>dep[p]+1&&e[i].w)

   {

      if (dep[e[i].v] > dep[p] + 1 && e[i].w) {

        dep[e[i].v] = dep[p] + 1;

    if(!vis[e[i].v])

    {

        if (!vis[e[i].v]) {

          vis[e[i].v] = 1;

          q.push(e[i].v);

        }
@@ -177,18 +160,15 @@ bool bfs() 
  if (dep[t] == INF) return 0;

  return 1;

}

int dfs(int p,int w)

{

int dfs(int p, int w) {

  if (p == t) return w;

  int used = 0;                           //已经使用的流量

  for (int i = cur[p]; i; i = e[i].next)  //每条边都尝试找一次增广路

  {

    cur[p] = i;  //当前弧优化

  if(dep[e[i].v]==dep[p]+1&&e[i].w)

  {

    if (dep[e[i].v] == dep[p] + 1 && e[i].w) {

      int flow = dfs(e[i].v, min(w - used, e[i].w));

   if(flow)

   {

      if (flow) {

        used += flow;

        e[i].w -= flow;

        e[i ^ 1].w += flow;
@@ -198,19 +178,16 @@ int dfs(int p,int w) 
  }

  return used;

}

int main()

{

int main() {

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

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

 {

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

    int u, v, w;

    scanf("%d%d%d", &u, &v, &w);

    addedge(u, v, w);

    addedge(v, u, 0);

  }

  int ans = 0;

 while(bfs())

  ans+=dfs(s,INF);

  while (bfs()) ans += dfs(s, INF);

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

  return 0;

}
@@ -219,4 +196,3 @@ int main() 
### ISAP



这个是 SAP 算法的加强版 (Improved)。