[WC2007]剪刀石头布
嘟嘟嘟
这题不愧是冬令营的题，有思维难度。
题面就是说有一个完全图，让你给一些无向边定向，使三元环最多。
这题关键就是怎么计数三元环。直接记非常难，所以我们要正难则反！三元环总数是\(C_{n} ^ {3}\)，然后考虑什么情况会破坏三元环：当一个点的出边大于1时，记\(d_i\)是\(i\)的出边数量，则破坏的三元环数量就是\(C_{d_i} ^ 2\)，所以三元环总数是\(C_{n} ^ {3} - \sum C_{d_i} ^ {2}\)。所以我们要最小化后面的那个东西。
观察\(C_{d_i} ^ {2} = \frac{d_i * (d_i - 1)}{2}\)，发现这东西其实是一个等比数列（这都能发现）。所以我们把一个点\(i\)拆成\(n\)个点，分别向汇点连容量为1，费用为\(,1, 2, 3 \ldots n - 1\)的边。
然后把边看成点。对于一条无向边\((x, y)\)，分别向点\(x, y\)连一条容量为1，费用为0的边，表示可以让其中一个点的出度+1；对于一条指向\(y\)的有向边，就只向点\(y\)连一条容量为1，费用为0的边。最后从源点向所有边代表的点连一条容量为1，费用为0的边。
至于输出矩阵，看对应边是否满流即可。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 105;
const int maxN = 2e4 + 5;
const int maxe = 1e7 + 5;
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
int n, t, a[maxn][maxn];
struct Edge
{
int nxt, from, to, cap, cos;
}e[maxe];
int head[maxN], ecnt = -1;
In void addEdge(int x, int y, int w, int c)
{
e[++ecnt] = (Edge){head[x], x, y, w, c};
head[x] = ecnt;
e[++ecnt] = (Edge){head[y], y, x, 0, -c};
head[y] = ecnt;
}
bool in[maxN];
int dis[maxN], pre[maxN], flow[maxN];
In bool spfa()
{
Mem(dis, 0x3f), Mem(in, 0);
dis[0] = 0, flow[0] = INF;
queue<int> q; q.push(0);
while(!q.empty())
{
int now = q.front(); q.pop(); in[now] = 0;
for(int i = head[now], v; ~i; i = e[i].nxt)
{
if(dis[v = e[i].to] > dis[now] + e[i].cos && e[i].cap > 0)
{
dis[v] = dis[now] + e[i].cos;
pre[v] = i;
flow[v] = min(flow[now], e[i].cap);
if(!in[v]) q.push(v), in[v] = 1;
}
}
}
return dis[t] ^ INF;
}
int minCost = 0;
In void update()
{
int x = t;
while(x)
{
int i = pre[x];
e[i].cap -= flow[t];
e[i ^ 1].cap += flow[t];
x = e[i].from;
}
minCost += flow[t] * dis[t];
}
In int MCMF()
{
minCost = 0;
while(spfa()) update();
return minCost;
}
In int N(int x, int y) {return n + (x - 1) * n + y;}
int ans[maxn][maxn];
In void solve(int x, int y)
{
for(int i = head[N(x, y)], v; ~i; i = e[i].nxt)
{
v = e[i].to;
if(v && v <= n && !e[i].cap)
{
if(v == y) ans[x][y] = 1;
else ans[y][x] = 1;
}
}
}
int main()
{
Mem(head, -1);
n = read(); t = n * n + n + 1;
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 = i + 1; j <= n; ++j)
{
addEdge(0, N(i, j), 1, 0);
if(a[i][j] == 2)
{
addEdge(N(i, j), i, 1, 0);
addEdge(N(i, j), j, 1, 0);
}
else if(a[i][j] == 1) addEdge(N(i, j), j, 1, 0);
else addEdge(N(i, j), i, 1, 0);
}
for(int i = 1; i <= n; ++i)
for(int j = 0; j < n; ++j) addEdge(i, t, 1, j);
write(1LL * n * (n - 1) * (n - 2) / 6 - MCMF()), enter;
for(int i = 1; i <= n; ++i)
for(int j = i + 1; j <= n; ++j) solve(i, j);
for(int i = 1; i <= n; ++i, enter)
for(int j = 1; j <= n; ++j) write(ans[i][j]), space;
return 0;
}
