2597: [Wc2007]剪刀石头布

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分析：　　

　　费用流。

　　首先转化一下问题，整张图最优的情况是存在$C_n^3$个，即任意3个都行，然后考虑去掉最少不满足的三元环。

　　如果u赢了v，u向v连一条边，如果v有k条入边，那么说明少了$C_k^2$个三元环，所对每场比赛分配度数，求最小费用最大流。

　　具体地：S向每场比赛连容量为1，花费为0的边；每场比赛向两个人连容量为1，花费为0的边；每个人因为度数不同，花费不同，所以差分后建边。

　　还有一种随机化+迭代的做法。

代码：

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<iostream>

#include<cmath>

#include<cctype>

#include<set>

#include<queue>

#include<vector>

#include<map>

using namespace std;

typedef long long LL;

inline int read() {

    int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;

    for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;

}

const int N = , INF = 1e9;

struct Edge { int from, to, nxt, cap, cost; } e[];

int head[N], pre[N], dis[N], q[], deg[N], id[][], A[][], P[N], tag[N], En = , S, T;

bool vis[N];

inline void add_edge(int u,int v,int f,int w) {

    ++En; e[En].from = u, e[En].to = v, e[En].cap = f, e[En].cost = w, e[En].nxt = head[u]; head[u] = En;

    ++En; e[En].from = v, e[En].to = u, e[En].cap = , e[En].cost = -w, e[En].nxt = head[v]; head[v] = En;

}

bool spfa() {

    for (int i = ; i <= T; ++i) dis[i] = INF, vis[i] = false, pre[i] = ;

    int L = , R = ;

    q[++R] = S, vis[S] = true, dis[S] = ;

    while (L <= R) {

        int u = q[L ++]; vis[u] = false;

        for (int i = head[u]; i; i = e[i].nxt) {

            int v = e[i].to;

            if (dis[v] > dis[u] + e[i].cost && e[i].cap > ) {

                dis[v] = dis[u] + e[i].cost;

                pre[v] = i;

                if (!vis[v]) q[++R] = v, vis[v] = true;

            }

        }

    }

    return dis[T] != INF;

}

int mcf() {

    int Cost = , Flow = ;

    while (spfa()) {

        int now = 1e9;

        for (int i = T; i != S; i = e[pre[i]].from)

            now = min(now, e[pre[i]].cap);

        for (int i = T; i != S; i = e[pre[i]].from)

            e[pre[i]].cap -= now, e[pre[i] ^ ].cap += now;

        Cost += dis[T] * now;

        Flow += now;

    }

    return Cost;

}

int main() {

    freopen("a.in", "r", stdin);

    int n = read(), tot = n;

    if (n < ) { printf(""); return ; }

    for (int i = ; i <= n; ++i) {

        for (int j = ; j <= i; ++j) read();

        for (int j = i + ; j <= n; ++j) {

            int x = read();

            id[i][j] = ++tot;

            if (x == ) deg[j] ++, tag[tot] = ;

            else if (x == ) deg[i] ++, tag[tot] = ;

            else { add_edge(tot, i, , ); P[tot] = En; add_edge(tot, j, , ); }

        }

    }

    S = , T = tot + ;

    for (int i = n + ; i <= tot; ++i) add_edge(S, i, , );

    for (int i = ; i <= n; ++i)

        for (int j = deg[i]; j < n - ; ++j) add_edge(i, T, , j);

    int ans = n * (n - ) * (n - ) / ;

    for (int i = ; i <= n; ++i)

        ans -= deg[i] * (deg[i] - ) / ;

    ans -= mcf();

    printf("%d\n", ans);

    for (int i = ; i <= n; ++i)

        for (int j = i + ; j <= n; ++j) {

            if (tag[id[i][j]]) A[i][j] = tag[id[i][j]] - ;

            else A[i][j] = e[P[id[i][j]]].cap ?  : ;

        }

    for (int i = ; i <= n; ++i, puts("")) {

        for (int j = ; j < i; ++j) printf("%d ", A[j][i] ^ );

        for (int j = i; j <= n; ++j) printf("%d ", A[i][j]);

    }

    return ;

}