BZOJ3571 & 洛谷3236：[HNOI2014]画框——题解

https://www.lydsy.com/JudgeOnline/problem.php?id=3571

https://www.luogu.org/problemnew/show/P3236

我能说我debug 1h才发现j打成i了吗TAT

对于最小代价二分图匹配可以KM算法做（当然费用流是可以的但是O（玄学）对吧……）。

这个所求表达式形式像极了最小乘积生成树：BZOJ2395：[Balkan 2011]Timeismoney

所以其实我们只是把生成树部分变成KM就行啦！

#include<cstdio>
#include<cctype>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=102;
const int INF=0x3f3f3f3f;
inline int read(){
int X=0,w=0;char ch=0;
while(!isdigit(ch)){w|=ch=='-';ch=getchar();}
while(isdigit(ch))X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
return w?-X:X;
}
struct point{
int x,y;
point(int xx=0,int yy=0){x=xx,y=yy;}
point operator -(const point &b)const{
return point(x-b.x,y-b.y);
}
int operator *(const point &b)const{
return x*b.y-y*b.x;
}
};
int n,ans,a

,b

;
int dis

,wx
,wy
,match
,sla
;
bool vx
,vy
;
bool dfs(int u){
vx[u]=1;
for(int v=1;v<=n;v++){
if(!vy[v]){
int w=wx[u]+wy[v]-dis[u][v];
if(!w){
vy[v]=1;
if(!match[v]||dfs(match[v])){
match[v]=u;return 1;
}
}else sla[v]=min(sla[v],w);
}
}
return 0;
}
point KM(){
point res=point(0,0);
memset(wx,0,sizeof(wx));
memset(wy,0,sizeof(wy));
memset(match,0,sizeof(match));
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
wx[i]=max(wx[i],dis[i][j]);
for(int i=1;i<=n;i++){
memset(sla,INF,sizeof(sla));
while(1){
memset(vx,0,sizeof(vx));
memset(vy,0,sizeof(vy));
if(dfs(i))break;
int minn=INF;
for(int j=1;j<=n;j++)
if(!vy[j])minn=min(minn,sla[j]);
for(int j=1;j<=n;j++){
if(vx[j])wx[j]-=minn;
if(vy[j])wy[j]+=minn;
else sla[j]-=minn;
}
}
}
for(int i=1;i<=n;i++)
res.x+=a[match[i]][i],res.y+=b[match[i]][i];
ans=min(ans,res.x*res.y);
return res;
}
void work(point l,point r){
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
dis[i][j]=-(a[i][j]*(l.y-r.y)+b[i][j]*(r.x-l.x));
point mid=KM();
if((r-l)*(mid-l)>=0)return;
work(l,mid);work(mid,r);
}
int main(){
int t=read();
while(t--){
ans=INF;n=read();
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
a[i][j]=read();
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
b[i][j]=read();
for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)dis[i][j]=-a[i][j];
point p1=KM();
for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)dis[i][j]=-b[i][j];
point p2=KM();
work(p1,p2);
printf("%d\n",ans);
}
return 0;
}

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+本文作者：luyouqi233。 　　　　　　　　　　　　　　+

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