Codeforces Round #381 (Div. 1) E - Gosha is hunting 最大费用网络流 期望

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
const int maxn=200010;
const double eps=1e-8;

#define sadf 123124
#define int long long
bool vis[maxn];
double n,m,x,y,z,f,cost[maxn],pre[maxn],flow[maxn],maxflow,mincost;
int s,t;
int last[maxn];
//cost最小花费;pre每个点的前驱；last每个点的所连的前一条边；flow源点到此处的流量
//maxflow 最大流量
//mincost 最大流量的情况下的最小花费
struct Edge
{
    int to,next;
    double flow,cost;//flow流量 cost花费
} edge[maxn];
int head[maxn],num_edge;
queue <int> q;
int aa,bb;
int AA,BB;
double qwq1[40010],qwq2[40010];
#define INF 2147483647
void add(int from,int to,int flow,double cost)
{
    edge[++num_edge].next=head[from];
    edge[num_edge].to=to;
    edge[num_edge].flow=flow;
    edge[num_edge].cost=cost;
    head[from]=num_edge;

    edge[++num_edge].next=head[to];
    edge[num_edge].to=from;
    edge[num_edge].flow=0;
    edge[num_edge].cost=-cost;
    head[to]=num_edge;
}
double spfa(int s,int t)
{
    for(int i=1; i<=t; ++i)
        cost[i]=-1;
    cost[s]=0;
    q.push(s);
    vis[s]=1;
    pre[t]=-1;
    flow[s]=2e9;//要加上这一条，不然到最后是flow都变为0  
    while (!q.empty())
    {
        int now=q.front();
        q.pop();
        vis[now]=0;
        for (int i=head[now]; i!=-1; i=edge[i].next)
        {
            if (edge[i].flow>eps && cost[edge[i].to]<cost[now]+edge[i].cost&&cost[now]+edge[i].cost-cost[edge[i].to]>eps)//正边
            {
                cost[edge[i].to]=cost[now]+edge[i].cost;
                pre[edge[i].to]=now;
                last[edge[i].to]=i;
                flow[edge[i].to]=min(flow[now],edge[i].flow);
                if (!vis[edge[i].to])
                {
                    vis[edge[i].to]=1;
                    q.push(edge[i].to);
                }
            }
        }
    }
    return pre[t]!=-1;
}

void MCMF()
{
    while (spfa(s,t))
    {
        int now=t;
        //回溯加的时候，要视情况，这个题，是把边的cost都加上
        //如果不把flow[s]初始化为正无穷，那么到t的时候，flow[t]就是0
        //若初始化为正无穷，就是1 
        maxflow+=flow[t]; 
        mincost+=flow[t]*cost[t];
        while (now!=s)
        {
//            mincost+=edge[last[now]].cost*edge[last[now]].flow;
            edge[last[now]].flow-=1;//flow和cost容易搞混
            edge[last[now]^1].flow+=1;
            now=pre[now];
        }
    }
}
signed main()
{
    memset(head,-1,sizeof(head));
    num_edge=1;
    cin>>n;
    cin>>aa>>bb;
    AA=n+1;
    BB=AA+1;
    s=BB+1;
    t=s+1;
    for(int i=1; i<=n; i++)
    {
        double x;
        cin>>x;
        qwq1[i]=x;
    }
    for(int i=1; i<=n; ++i)
    {
        double x;
        cin>>x;
        qwq2[i]=x;
    }
    add(s,AA,aa,0.0);
    add(s,BB,bb,0.0);
    for(int i=1; i<=n; i++)
    {
        add(AA,i,1,qwq1[i]);
        add(BB,i,1,qwq2[i]);
        add(i,t,1,0.0);
        add(i,t,1,-(qwq1[i]*qwq2[i]));
    }
    MCMF();
    printf("%.8lf\n",mincost);
    return 0;
}