【BZOJ 4819】【SDOI 2017】新生舞会

一看到这个要求的式子就能想到01分数规划，二分答案x后任意两个人对答案的贡献是a[i][j]-xb[i][j]。这样问题就转化为一个二分图匹配，边权就是这个贡献。
显然这是一个裸的费用流，首先建立超级源点和超级汇点，然后所有边的流量上限都是1，男生的n个点和女生的n个点之间两两连边，费用单价就是对答案贡献取反（因为求的是最大费用），加反向边，跑一下费用流就好了。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#define ll long long
#define inf 20000000.0
#define mod 1000000007
#define N 40000
#define M 205
#define eps 1e-7
#define fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
using namespace std;
queue<int> q;
int to[N],sz[N],nxt[N],head[M],vis[N],list[N],fa[N];
int tot,S,T,n,i,j;
double cost,l,r,res,mid,ct[N],dis[N];
int a[M][M],b[M][M];
void ADD(int x,int y,int u,double cost)
{to[++tot]=y; sz[tot]=u; ct[tot] = cost; nxt[tot] = head[x]; head[x] = tot;}
void add(int x,int y,double z) {ADD(x,y,1,z); ADD(y,x,0,-z);}
bool spfa()
{
    int i,h,t,nt,nw;
    fo(i,S,T) dis[i] = inf , vis[i] = 0;
    vis[S] = 1; dis[S] = 0;
    q.push(S);
    while (!q.empty())
        {
            nw = q.front(); vis[nw] = 0; q.pop();
            for (i = head[nw]; i ; i = nxt[i])
                {
                    nt = to[i];
                    if (sz[i] && (dis[nt] - (dis[nw]+ct[i]) >= eps))
                        {
                            dis[nt] = dis[nw] + ct[i];
                            fa[nt] = i;
                            if (!vis[nt]) {q.push(nt); vis[nt] = 1;}
                        }
                }
        }
    if (dis[T] == inf) return false;
    int f = inf;
    for (i = fa[T]; i ; i = fa[to[i^1]]) f = min(f,sz[i]);
    for (i = fa[T]; i ; i = fa[to[i^1]]) sz[i] -= f , sz[i^1] += f;
    cost += dis[T] * f;
    return true;
}

bool judge(double x)
{
    int i,j;
    tot = 1; memset(head,0,sizeof(head));
    fo(i,1,n)
        fo(j,1,n)
            add(i,j+n,-(a[i][j] - 1.0 * x * b[i][j]));
    fo(i,1,n) add(S,i,0) , add(i+n,T,0);
    cost = 0;  while (spfa()); cost = -cost;
    return cost - 0 >= eps;
}
int main()
{
    scanf("%d",&n); S = 0; T = 2 * n + 1;
    fo(i,1,n) fo(j,1,n) scanf("%d",&a[i][j]);
    fo(i,1,n) fo(j,1,n) scanf("%d",&b[i][j]);
    l = 0.0; r = 10000.0; res = 0.0;
    while (r - l >= eps)
        {
            mid = (l + r) / 2;
            if (judge(mid)) l = mid , res = mid; else r = mid;  
        }
    printf("%.6lf",res);
    return 0;
}