OpenJPOJ1013_Excellent Note_Dinic算法求二分图匹配

题意

求给出正则二分图的最大匹配。

思路

匈牙利算法果断TLE，用dinic算法求解二分图匹配。

算法过程

1.读入正则二分图，dfs黑白染色

2.在染色图上建立跑dinic的图。s到白点，白点到黑点，黑点到t，权值都是1。这里存图有一定技巧，在方便地找反向边地同时，还能保证原边的位号都是偶数，方便最后输出结果。

3.dinic算法，输出结果。

链接

https://vjudge.net/contest/179392#problem/D

代码

#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>

using namespace std;

const int maxn = 1e5 + 10;
const int maxm = 1e6 + 10;
const int maxe = maxn * 2 + maxm;
const int inf  = 0x3f3f3f3f;

struct edge{
int to, cap, next;
};

int n, m;
int x, y;

vector<int> G[maxn];
int mark[maxn];

int head[maxn], tt;
edge E[maxe];
int level[maxn], iter[maxn];

void dfs0(int v, int c)
{
mark[v] = c;
for(int i = 0; i < G[v].size(); i++)
{
int d = G[v][i];
if(mark[d] < 0)
{
dfs0(d, c ^ 1);
}
}
}

void get_color()
{
memset(mark, -1, sizeof mark);
for(int i = 1; i <= n; i++)
{
if(mark[i] >= 0) continue;
dfs0(i, 0);
}
}

void add(int from, int to, int cap)
{
E[tt].to = to, E[tt].cap = cap, E[tt].next = head[from], head[from] = tt, tt++;
}

void build_g(int s, int t)
{
tt = 0;
memset(
4000
head, -1, sizeof head);
for(int i = 1; i <= n; i++)
{
if(mark[i] == 0)
{
add(s, i, 1);
add(i, s, 0);

for(int j = 0; j < G[i].size(); j++)
{
int d = G[i][j];
add(i, d, 1);
add(d, i, 0);
}
}
else{
add(i, t, 1);
add(t, i, 0);
}
}
}

void bfs(int s)
{
memset(level, -1, sizeof level);
queue<int> qu;

level[s] = 0;
qu.push(s);

while(qu.size())
{
int v = qu.front();
qu.pop();

for(int i = head[v]; i != -1; i = E[i].next)
{
edge e = E[i];
if(e.cap > 0 && level[e.to] < 0)
{
level[e.to] = level[v] + 1;
qu.push(e.to);
}
}
}
}

int dfs(int v, int t, int f)
{
if(v == t) return f;
for(int &i = iter[v]; i != -1; i = E[i].next)
{
edge &e = E[i];
if(e.cap > 0 && level[e.to] > level[v])
{
int d = dfs(e.to, t, min(f, e.cap));
if(d > 0)
{
e.cap -= d;
E[i^1].cap += d;
return d;
}
}
}

return 0;
}

int dinic(int s, int t)
{
int flow = 0;
while(1)
{
bfs(s);
if(level[t] < 0) return flow;

for(int i = s; i <= t; i++) iter[i] = head[i];
while(1)
{
int f = dfs(s, t, inf);
if(f == 0) break;
flow += f;
}
}
}

void output()
{
for(int i = 0; i < tt; i += 2)
{
if(E[i].cap == 0)
{
if(E[i].to == 0 || E[i].to == n+1) continue;
if(E[i^1].to == 0 || E[i^1].to == n+1) continue;
cout << E[i].to << " " << E[i^1].to << endl;
}
}
}

int main()
{
scanf("%d %d", &n, &m);

for(int i = 0; i < m; i++)
{
scanf("%d %d", &x, &y);
G[x].push_back(y);
G[y].push_back(x);
}

get_color();

int s = 0, t = n + 1;
build_g(s, t);

dinic(s, t);

output();

return 0;
}