标签:size def solution print void += cst ack names
找到和最小的两个序列a,b满足对于任意i,j有a[i]+b[j]>=c[i][j](矩阵c给出)。
裸的二分图就水过了……
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define clr(a,m) memset(a,m,sizeof(a))
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int MAXN=555;
const int INF =1e9;
const double eps=1e-10;
int gap[MAXN][MAXN];
int Lx[MAXN],Ly[MAXN],slack[MAXN];
int left[MAXN],n;
bool S[MAXN],T[MAXN];
void read()
{
rep(i,1,n)
rep(j,1,n)
scanf("%d",&gap[i][j]);
}
bool match(int u)
{
S[u]=true;
rep(v,1,n)
if(!T[v]){
int tmp=Lx[u]+Ly[v]-gap[u][v];
if(tmp==0){
T[v]=true;
if(!left[v]||match(left[v])){
left[v]=u;
return true;
}
}else slack[v]=min(slack[v],tmp);
}
return false;
}
void update()
{
int a=INF;
rep(v,1,n)
if(!T[v])
a=min(a,slack[v]);
rep(i,1,n){
if(S[i])Lx[i]-=a;
if(T[i])Ly[i]+=a;
}
}
void KM()
{
rep(i,1,n){
left[i]=Ly[i]=0;
Lx[i]=-INF;
rep(j,1,n)
Lx[i]=max(Lx[i],gap[i][j]);
}
rep(i,1,n){
rep(j,1,n)
slack[j]=INF;
while(1)
{
rep(j,1,n)
S[j]=T[j]=0;
if(match(i))
break;
else
update();
}
}
}
void print()
{
int ans=0;
rep(i,1,n){
ans+=Lx[i];
if(i!=1)printf(" ");
printf("%d",Lx[i]);
}
printf("\n");
rep(i,1,n){
ans+=Ly[i];
if(i!=1)printf(" ");
printf("%d",Ly[i]);
}
printf("\n");
printf("%d\n",ans);
}
int main()
{
while(~scanf("%d",&n))
{
read();
KM();
print();
}
return 0;
}
标签:size def solution print void += cst ack names
原文地址:https://www.cnblogs.com/rui-4825/p/12870759.html