题解#3

 1 int n,m,k,mincost,tot=1,cnt,s,t,win[maxn],a[maxn][maxn],b[maxn][maxn],head[maxm],d[maxm],from[2*maxm];
 2   
 3 bool v[maxm];
 4   
 5 queue<int>q;
 6   
 7 struct edge{int from,go,next,v,c;}e[2*maxm];
 8   
 9 void add(int x,int y,int v,int c)
10   
11 {
12   
13     e[++tot]=(edge){x,y,head[x],v,c};head[x]=tot;
14   
15     e[++tot]=(edge){y,x,head[y],0,-c};head[y]=tot;
16   
17 }
18   
19 bool spfa()
20   
21 {
22   
23     for (int i=0;i<=cnt;i++){v[i]=0;d[i]=inf;}
24   
25     q.push(s);d[s]=0;v[s]=1;
26   
27     while(!q.empty())
28   
29     {
30   
31         int x=q.front();q.pop();v[x]=0;
32   
33         for (int i=head[x],y;i;i=e[i].next)
34   
35          if(e[i].v&&d[x]+e[i].c<d[y=e[i].go])
36   
37          {
38   
39             d[y]=d[x]+e[i].c;from[y]=i;
40   
41             if(!v[y]){v[y]=1;q.push(y);}
42   
43          }
44   
45     }
46   
47     return d[t]!=inf;
48   
49 }
50   
51 void mcf()
52   
53 {
54   
55     while(spfa())
56   
57     {
58   
59         int tmp=inf;
60   
61         for(int i=from[t];i;i=from[e[i].from]) tmp=min(tmp,e[i].v);
62   
63         mincost+=d[t]*tmp;
64   
65         for(int i=from[t];i;i=from[e[i].from]){e[i].v-=tmp;e[i^1].v+=tmp;}
66   
67     }
68   
69 }
70   
71 int main()
72   
73 {
74     n=read();s=0;t=n+1;cnt=t;
75     for5(n,n)a[i][j]=read(),win[i]+=(a[i][j]==1);
76     for1(i,n)for1(j,i-1)if(a[i][j]==2)
77     {
78         add(s,++cnt,1,0);
79         b[i][j]=tot+1;
80         add(cnt,i,1,0);
81         add(cnt,j,1,0);
82     }
83     for1(i,n)
84     {
85         mincost+=win[i]*(win[i]-1)/2;
86         for2(j,win[i],n-2)add(i,t,1,j);
87     }
88     mcf();
89     cout<<(n*(n-1)*(n-2)/6-mincost)<<endl;
90     for1(i,n)for1(j,i-1)if(a[i][j]==2)a[i][j]=!e[b[i][j]].v,a[j][i]=!a[i][j];
91     for1(i,n)for1(j,n)printf("%d%c",a[i][j],j==n?‘\n‘:‘ ‘);
92   
93     return 0;
94   
95 }