P3705 [SDOI2017]新生舞会

题意

看见这个式子就知道应该\(0-1\)分数规划了。

从\(S\)向每个男生连容量为\(1\)费用为\(0\)的边，从没个女生向\(T\)连容量为\(1\)费用为\(0\)的边。

二分答案\(mid\)。

对于点对\((i,j)\)，\(i\)是男生，\(j\)是女生，从\(i\)向\(j+n\)连容量为\(1\)费用为\(a_{i,j}-mid*b_{i,j}\)的边。

之后跑最大费用最大流，可行则费用大于等于\(0\)。

code:

#include<bits/stdc++.h>
using namespace std;
const int maxn=210;
const double inf=1000000000;
const double eps=1e-8;
int n,cnt_edge=1,S,T;
int head[maxn];
double sum;
double dis[maxn];
double a[maxn][maxn],b[maxn][maxn];
bool vis[maxn];
struct edge{int to,nxt,flow;double cost;}e[maxn*maxn*4];
inline void add(int u,int v,int w,double c)
{
    e[++cnt_edge].nxt=head[u];
    head[u]=cnt_edge;
    e[cnt_edge].to=v;
    e[cnt_edge].flow=w;
    e[cnt_edge].cost=c;
}
inline void addflow(int u,int v,int w,double c){add(u,v,w,c);add(v,u,0,-c);}
inline bool spfa()
{
    memset(vis,0,sizeof(vis));
    for(int i=S;i<=T;i++)dis[i]=-inf;
    queue<int>q;
    q.push(S);dis[S]=0;vis[S]=1;
    while(!q.empty())
    {
        int x=q.front();q.pop();vis[x]=0;
        for(int i=head[x];i;i=e[i].nxt)
        {
            if(e[i].flow<=0)continue;
            int y=e[i].to;
            if(dis[y]<dis[x]+e[i].cost)
            {
                dis[y]=dis[x]+e[i].cost;
                if(!vis[y])q.push(y),vis[y]=1;
            }
        }
    }
    return dis[T]!=-inf;
}
int dfs(int x,int lim)
{
    vis[x]=1;
    if(lim<=0||x==T)return lim;
    int res=lim;
    for(int i=head[x];i;i=e[i].nxt)
    {
        int y=e[i].to;
        if(e[i].flow<=0||vis[y]||dis[y]!=dis[x]+e[i].cost)continue;
        int tmp=dfs(y,min(res,e[i].flow));
        res-=tmp;
        e[i].flow-=tmp,e[i^1].flow+=tmp;
        if(res<=0)break;
    }
    return lim-res;
}
inline double Dinic()
{
    int res=0;double cost=0;
    while(spfa())
    {
        int flow=dfs(S,inf);
        res+=flow,cost+=1.0*dis[T]*flow;
    }
    return cost;
}
inline bool check(double mid)
{
    memset(head,0,sizeof(head));
    cnt_edge=1;
    S=0,T=2*n+1;
    for(int i=1;i<=n;i++)addflow(S,i,1,0),addflow(i+n,T,1,0);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            addflow(i,j+n,1,a[i][j]-mid*b[i][j]);
    return Dinic()>=0;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            scanf("%lf",&a[i][j]),sum+=a[i][j];
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            scanf("%lf",&b[i][j]);
    double l=0,r=sum,ans=0;
    while(r-l>eps)
    {
        double mid=(l+r)/2.0;
        if(check(mid))ans=mid,l=mid+eps;
        else r=mid-eps;
    }
    printf("%.6lf",ans);
    return 0;
}