Zero' Site

二分图匈牙利算法，归根结底就是递归找新边找找找！

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<stack>
#include<cmath>
#include<ctime>
#include<queue>
#include<deque>
#include<list>
#include<map>
#define int long long
#define ffor(i, a, b) for(ll i = a; i <= b; i++)
#define rfor(i, a, b) for(ll i = a; i >= b; i--)
#define mes(a,b) memset(a, b, sizeof(a))
#define cos(x) cos(x*PI/180.0)
#define sin(x) sin(x*PI/180.0)
#define stop system("pause")
#define see(s,x) cout<<(s)<<'='<<(x)<<endl
#define IMAX 0x7fffffff
#define PI 3.141592654
#define INF 0x3f3f3f3f
#define eps 1e-6
#define lowbit(x) (x&(-x))
#define test(x) cout<<"Test "<<(x)<<endl;
typedef long long ll;
ll mod;
ll max(ll a, ll b) { return(a > b) ? a : b; }
ll min(ll a, ll b) { return(a < b) ? a : b; }
using namespace std;

vector<int> G[1005];
int tab[1005];
bool v[1005];

bool deal(int x)
{
    for (int i : G[x])
    {
        if (!v[i])
        {
            v[i] = true;
            if (!tab[i] || deal(tab[i]))
            {
                tab[i] = x;
                return true;
            }
        }
    }
    return false;
}

signed main()
{
    mes(tab, 0);
    int n, m, e;
    cin >> n >> m >> e;
    ffor(i, 1, e)
    {
        int x, y;
        scanf("%lld%lld", &x, &y);
        if (x <= n && y <= m)
            G[x].push_back(y);
    }
    int res = 0;
    ffor(i, 1, n)
    {
        mes(v, 0);
        res += deal(i);
    }
    cout << res << endl;
}

输入格式

输出格式

输入输出样例

4 5 4 3
4 2 30 2
4 3 20 3
2 3 20 1
2 1 30 9
1 3 40 5

50 280

说明/提示

思路：初学费用流，找到了一篇不错的教程，思路是利用最短路保证最小费用，再在最大流的基础上，算出最大流的最小的费用，似乎可以把sfpa换成dijkstra？

AC代码：

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<stack>
#include<cmath>
#include<ctime>
#include<queue>
#include<deque>
#include<list>
#include<map>
#define int long long
#define ffor(i, a, b) for(ll i = a; i <= b; i++)
#define rfor(i, a, b) for(ll i = a; i >= b; i--)
#define mes(a,b) memset(a, b, sizeof(a))
#define cos(x) cos(x*PI/180.0)
#define sin(x) sin(x*PI/180.0)
#define stop system("pause")
#define see(s,x) cout<<(s)<<'='<<(x)<<endl
#define IMAX 0x7fffffff
#define PI 3.141592654
#define INF 0x3f3f3f3f
#define eps 1e-6
#define lowbit(x) (x&(-x))
#define test(x) cout<<"Test "<<(x)<<endl;
typedef long long ll;
ll mod;
ll max(ll a, ll b) { return(a > b) ? a : b; }
ll min(ll a, ll b) { return(a < b) ? a : b; }
using namespace std;

struct unit
{
    int to;//指向
    int f;//流量
    int w;//单位花费
    int rev;//反向边序号
    unit() {};
    unit(int n, int ff, int ww, int r) :to(n), f(ff), w(ww), rev(r) {};
};
vector<unit> G[50001];
int cur[50001];
int dis[50001];
bool v[50001];

int n, m, st, ed;
int maxflow, mincost;

void add(int x, int y, int f, int w)//建边
{
    G[x].push_back(unit(y, f, w, G[y].size()));
    G[y].push_back(unit(x, 0, -w, G[x].size() - 1));//反向边，流量为0，单位花费为负
}

bool spfa()//标准SPFA
{
    ffor(i, 1, n)
    {
        v[i] = false;
        dis[i] = IMAX;
    }
    dis[st] = 0;
    queue<int> q;
    q.push(st);
    while (!q.empty())
    {
        int f = q.front();
        q.pop();
        v[f] = false;
        for (vector<unit>::iterator it = G[f].begin(); it != G[f].end(); it++)
        {
            unit u = *it;
            if (u.f && dis[f] + u.w < dis[u.to])
            {
                dis[u.to] = dis[f] + u.w;
                if (!v[u.to])
                {
                    q.push(u.to);
                    v[u.to] = true;
                }
            }
        }
    }
    return dis[ed] != IMAX;//能走到就返回
}

int dfs(int x, int f)
{
    if (x == ed)
        return f;
    v[x] = true;//防止重复走
    int sum = 0;
    for (int& i = cur[x]; i <= G[x].size() - 1; i++)
    {
        unit& u = G[x][i];
        if (!v[u.to] && u.f && dis[u.to] == dis[x] + u.w)
        {
            int flow = dfs(u.to, min(f - sum, u.f));
            if (flow)//确认可以流通，就对边进行操作，记录最大流和最小花费
            {
                sum += flow;
                u.f -= flow;
                G[u.to][u.rev].f += flow;
                mincost += u.w * flow;
                if (sum == f)break;
            }
        }
    }
    v[x] = false;
    return sum;
}
void dinic()
{
    while (spfa())
    {
        mes(cur, 0);
        maxflow += dfs(st, IMAX);
    }
}

signed main()
{
    cin >> n >> m >> st >> ed;
    maxflow = mincost = 0;
    ffor(i, 1, m)
    {
        int x, y, w, c;
        scanf("%lld%lld%lld%lld", &x, &y, &w, &c);
        add(x, y, w, c);
    }
    dinic();
    printf("%lld %lld\n", maxflow, mincost);
}