【bzoj3698】XWW的难题 有上下界的网络流

源点S向每一行连一条容量为(a[i]
,a[i]
+1)的边

每一列向汇点T连一条容量为(a
[i],a
[i]+1)的边

行i向列j连一条容量为(a[i][j],a[i][j]+1)的边

求最大流

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#define maxn 210
#define maxm 100010
#define inf 1000000000

using namespace std;

int head[maxn],to[maxm],c[maxm],next[maxm],q[maxn],d[maxn],in[maxn],cnt[maxn];
double a[maxn][maxn];
int n,m,num,s,t,S,T,ans;

void addedge(int x,int y,int z)
{
num++;to[num]=y;c[num]=z;next[num]=head[x];head[x]=num;
num++;to[num]=x;c[num]=0;next[num]=head[y];head[y]=num;
}

bool bfs()
{
memset(d,-1,sizeof(d));
int l=0,r=1;
q[1]=S;d[S]=0;
while (l<r)
{
int x=q[++l];
for (int p=head[x];p;p=next[p])
if (c[p] && d[to[p]]==-1)
{
d[to[p]]=d[x]+1;
q[++r]=to[p];
}
}
if (d[T]==-1) return 0; else return 1;
}

int find(int x,int low)
{
if (x==T || low==0) return low;
int totflow=0;
for (int p=head[x];p;p=next[p])
if (c[p] && d[to[p]]==d[x]+1)
{
int a=find(to[p],min(c[p],low));
c[p]-=a;c[p^1]+=a;
low-=a;totflow+=a;
if (low==0) return totflow;
}
if (low) d[x]=-1;
return totflow;
}

void Dinic()
{
int ans=0;
while (bfs()) ans+=find(S,inf);
}

bool check(int x)
{
num=1;s=0;t=2*n+1;S=t+1;T=S+1;
memset(head,0,sizeof(head));
memset(in,0,sizeof(in));
memset(cnt,0,sizeof(cnt));
in[s]+=x;in[t]-=x;addedge(t,s,inf);
for (int i=1;i<n;i++)
{
if (a[i]
!=(int)a[i]
) addedge(s,i,1);
in[s]-=(int)a[i]
;in[i]+=(int)a[i]
;
}
for (int i=1;i<n;i++)
{
if (a
[i]!=(int)a
[i]) addedge(i+n,t,1);
in[i+n]-=(int)a
[i];in[t]+=(int)a
[i];
}
for (int i=1;i<n;i++)
for (int j=1;j<n;j++)
{
if (a[i][j]!=(int)a[i][j]) addedge(i,j+n,1);
in[i]-=(int)a[i][j];in[j+n]+=(int)a[i][j];
}
for (int i=s;i<=t;i++)
if (in[i]>0) cnt[i]=num+1,addedge(S,i,in[i]);
else if (in[i]<0) cnt[i]=num+1,addedge(i,T,-in[i]);
Dinic();
for (int i=s;i<=t;i++)
if (cnt[i] && c[cnt[i]]) return 0;
return 1;
}

int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
scanf("%lf",&a[i][j]);
int l=0,r=inf,ans=-1;
while (l<=r)
{
int mid=(l+r)/2;
if (check(mid)) ans=mid,l=mid+1; else r=mid-1;
}
if (ans==-1) printf("No\n");
else printf("%d\n",ans*3);
return 0;
}