luoguP3705 [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;
}