最小费用最大流( 洛谷3381)

版权声明:写了自己看的,看不懂不能怪我emmmm。 https://blog.csdn.net/qq_40858062/article/details/82952085

题目链接:https://www.luogu.org/problemnew/show/P3381

思路:模板题,改都不带改的

#pragma GCC optimize(2)
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(6666666)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
const int N=1e4+7;
struct edge{int to,cap,cost,rev;};
vector<edge>g[N];
int h[N],d[N],pre[N],prv[N],n,m,res,s,t,flow;
void add(int from,int to,int cap,int cost)
{
    g[from].pb(edge{to,cap,cost,(int)g[to].size()});
    g[to].pb(edge{from,0,-cost,(int)g[from].size()-1});
}
void slove(int s,int t,int f,int n)
{
    //int res=0;
    fill(h,h+n+1,0);
    while(f>0)
    {
        priority_queue<P,vector<P>,greater<P> > q;
        fill(d,d+n+1,inf);
        d[s]=0;
        q.push(P(0,s));
        while(!q.empty())
        {
            P p=q.top();q.pop();
            int v=p.second;
            if(d[v]<p.first) continue;
            for(int i=0;i<(int)g[v].size();i++)
            {
                edge &e=g[v][i];
                if(e.cap>0&&d[e.to]>d[v]+e.cost+h[v]-h[e.to])
                {
                    d[e.to]=d[v]+e.cost+h[v]-h[e.to];
                    prv[e.to]=v;pre[e.to]=i;
                    q.push(P(d[e.to],e.to));
                }
            }
        }
        if(d[t]==inf) break;
        FOR(i,0,n) h[i]+=d[i];
        int d=f;
        for(int i=t;i!=s;i=prv[i])  d=min(d,g[prv[i]][pre[i]].cap);
        f-=d,flow+=d;
        res+=d*h[t];
        for(int i=t;i!=s;i=prv[i])
        {
            edge &e=g[prv[i]][pre[i]];
            e.cap-=d;
            g[i][e.rev].cap+=d;
        }
    }
    //return res;
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>n>>m>>s>>t;
    int x,y,z,q;
    FOR(i,1,m)
    {
        si(x),si(y),si(z),si(q);
        add(x,y,z,q);
    }
    slove(s,t,inf,n);
    cout<<flow<<" "<<res<<endl;
    return 0;
}

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转载自blog.csdn.net/qq_40858062/article/details/82952085