题目链接
题解
orz思维差
既然是一张竞赛图，我们选出任意三个点都可能成环
总方案数为
\[{n \choose 3}\]
如果三个点不成环，会发现它们的度数是确定的，入度分别为\(2,1,0\)，出度为\(0,1,2\)
所以一个点的任意两个入度，都会对答案产生一个负的贡献
所以三元环数量为
\[{n \choose 3} - \sum\limits_{i = 1}^{n} {inde[i] \choose 2}\]
我们要最大化三元环数目，就要最小化\(\sum\limits_{i = 1}^{n} {inde[i] \choose 2}\)
考虑建模，使用费用流
每条边可以将入度分给，也仅可以分配给两端点中的一个
我们就每条边建一个点，从\(S\)向每条边建出来的点连一条\((1,0)\)的边，表示能产生一个流量
然后该边的点向那两个端点分别连一条\((1,0)\)的边，表示能产生\(1\)个入度
然后考虑每产生一个入度的影响
考虑到
\[{x \choose 2} - {x - 1 \choose 2} = x - 1\]
所以每增加一个入度，使得入度为\(x\)时，会多产生\(x - 1\)个贡献
按照费用流的套路，我们对每个点每一种度数建一条到\(T\)的边(1,x - 1)，表示消耗这么多三元环
按照费用流的性质，一定会优先选择权值较小的边，也就是逐层增加
建图时，还要考虑原来已有的边
然后就做完了
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 11005,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int h[maxn],ne = 1;
struct EDGE{int from,to,nxt,f,w;}ed[maxm];
inline void build(int u,int v,int f,int w){
ed[++ne] = (EDGE){u,v,h[u],f,w}; h[u] = ne;
ed[++ne] = (EDGE){v,u,h[v],0,-w}; h[v] = ne;
}
int q[maxn * 10],head,tail;
int p[maxn],vis[maxn],S,T;
int d[maxn],minf[maxn];
int mincost(){
int cost = 0,flow = 0;
while (true){
for (int i = S; i <= T; i++) d[i] = minf[i] = INF,vis[i] = false;
d[S] = 0; q[head = tail = 0] = S;
int u;
while (head <= tail){
u = q[head++];
vis[u] = false;
Redge(u) if (ed[k].f && d[u] + ed[k].w < d[to = ed[k].to]){
minf[to] = min(minf[u],ed[k].f);
p[to] = k; d[to] = d[u] + ed[k].w;
if (!vis[to]) q[++tail] = to,vis[to] = true;
}
}
if (d[T] == INF) break;
flow += minf[T]; cost += d[T] * minf[T];
u = T;
while (u != S){
ed[p[u]].f -= minf[T];
ed[p[u] ^ 1].f += minf[T];
u = ed[p[u]].from;
}
}
return cost;
}
int n,G[105][105],de[105],N,a[maxn],b[maxn];
int main(){
n = read();
REP(i,n) REP(j,n){
G[i][j] = read();
if (i == j) continue;
if (!G[i][j]) de[i]++;
else if (G[i][j] == 2 && i < j) N++,a[N] = i,b[N] = j;
}
S = 0; T = N + n + 1;
REP(i,N) build(S,i,1,0),build(i,N + a[i],1,0),build(i,N + b[i],1,0);
REP(i,n){
for (int j = de[i] + 1; j <= n; j++)
build(N + i,T,1,j - 1);
}
int ans = n * (n - 1) * (n - 2) / 6;
REP(i,n) if (de[i] > 1) ans -= de[i] * (de[i] - 1) / 2;
ans -= mincost();
printf("%d\n",ans);
REP(i,N){
Redge(i) if ((to = ed[k].to) > N && !ed[k].f){
if (to - N == a[i]) G[a[i]][b[i]] = 0,G[b[i]][a[i]] = 1;
else G[a[i]][b[i]] = 1,G[b[i]][a[i]] = 0;
break;
}
}
for (int i = 1; i <= n; i++,puts("")){
for (int j = 1; j <= n; j++){
printf("%d",G[i][j]);
if (j < n) putchar(' ');
}
}
return 0;
}