BZOJ 3571 [Hnoi2014]画框（最小乘积完美匹配）

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
const int N=310;
const LL INF=0x3f3f3f3f3f3f3f3f;
int nx,ny; //两边的点数 
LL g[N][N];  //二分图描述
int linker[N];//y中各点匹配状态
LL lx[N],ly[N];//x,y中的点标号
int n;
LL slack[N];
bool visx[N],visy[N];
LL ans=INF;
int T,a[N][N],b[N][N];
struct P{
    int x,y;
    P(){x=y=0;}
    P(int _x,int _y){x=_x;y=_y;}
    P operator-(const P&rhs){return P(x-rhs.x,y-rhs.y);}
}l,r;
LL cross(P a,P b){return (LL)a.x*b.y-(LL)a.y*b.x;}
bool DFS(int x){
    visx[x]=1;
    for(int y=0;y<ny;y++){
        if(visy[y])continue;
        int tmp=lx[x]+ly[y]-g[x][y];
        if(tmp==0){
            visy[y]=true;
            if(linker[y]==-1||DFS(linker[y])){
                linker[y]=x;
                return 1;
            }
        }else if(slack[y]>tmp)slack[y]=tmp;
    }return 0;
}
P KM(){
    P p;
    memset(linker,-1,sizeof(linker));
    memset(ly,0,sizeof(ly));
    for(int i=0;i<nx;i++){
        lx[i]=-INF;
        for(int j=0;j<ny;j++)if(g[i][j]>lx[i])lx[i]=g[i][j];
    }
    for(int x=0;x<nx;x++){
        for(int i=0;i<ny;i++)slack[i]=INF;
        for(;;){
            memset(visx,false,sizeof(visx));
            memset(visy,false,sizeof(visy));
            if(DFS(x))break;
            LL d=INF;
            for(int i=0;i<ny;i++)if(!visy[i]&&d>slack[i])d=slack[i];
            for(int i=0;i<nx;i++)if(visx[i])lx[i]-=d;
            for(int i=0;i<ny;i++){
                if(visy[i])ly[i]+=d;
                else slack[i]-=d;
            }
        }
    }LL res=0;
    for(int i=0;i<ny;i++)if(linker[i]!=-1){
        p.x+=a[linker[i]][i];
        p.y+=b[linker[i]][i];
    }res=(LL)p.x*p.y;
    if(res<ans)ans=res; 
    return p;
}
void work(P l,P r){
    P t=l-r;
    for(int i=0;i<n;i++)for(int j=0;j<n;j++)g[i][j]=-cross(P(a[i][j],b[i][j]),t);
    P mid=KM();
    if(cross(mid-l,r-mid)>0)work(l,mid),work(mid,r);
}
int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n); ans=INF;
        for(int i=0;i<n;i++)for(int j=0;j<n;j++)scanf("%d",&a[i][j]);
        for(int i=0;i<n;i++)for(int j=0;j<n;j++)scanf("%d",&b[i][j]);
        nx=ny=n;
        for(int i=0;i<n;i++)for(int j=0;j<n;j++)g[i][j]=-a[i][j]; l=KM();
        for(int i=0;i<n;i++)for(int j=0;j<n;j++)g[i][j]=-b[i][j]; r=KM();
        work(l,r); 
        printf("%lld\n",ans);
    }return 0;
}