Luogu4249 WC2007 石头剪刀布 费用流

传送门

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考虑竞赛图三元环计数，设第\(i\)个点的入度为\(d_i\)，根据容斥，答案为\(C_n^3 - \sum C_{d_i}^2\)

所以我们需要最小化\(\sum C_{d_i}^2\)

考虑将\(C_{d_i}^2\)差分，然后通过费用流解决

下面\((a,b)\)边表示流量为\(a\)、费用为\(b\)的边

建图：

每一场比赛和每一个人都建一个点

\(S\)向每一场比赛连\((1,0)\)边

每一场比赛若不确定结果则向两个参与者连\((1,0)\)边，如果胜者确定则只向胜者连\((1,0)\)边

每一个选手向\(T\)连\((1,C_i^2-C_{i-1}^2)(i \in [1,N])\)边

因为\(C_i^2-C_{i-1}^2=i-1\)，所以\(j>i \rightarrow C_j^2 - C_{j-1}^2 > C_i^2 - C_{i-1}^2\)，所以每一次一场比赛确定胜者能够正确地增加TA的贡献。

跑最小费用最大流，最后的答案就是最小的\(\sum C_{d_i}^2\)

输出方案考虑每一场比赛对应的点的流流向哪一个选手即可

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
//This code is written by Itst
using namespace std;

inline int read(){
    int a = 0;
    char c = getchar();
    bool f = 0;
    while(!isdigit(c) && c != EOF){
        if(c == '-')
            f = 1;
        c = getchar();
    }
    if(c == EOF)
        exit(0);
    while(isdigit(c)){
        a = (a << 3) + (a << 1) + (c ^ '0');
        c = getchar();
    }
    return f ? -a : a;
}

const int MAXN = 20010 , MAXM = 500010;
struct Edge{
    int end , upEd , f , c;
}Ed[MAXM];
int head[MAXN] , dis[MAXN] , pre[MAXN] , flo[MAXN] , ans[110][110];
int N , S , T , cntEd = 1;
bool vis[MAXN];
queue < int > q;

inline void addEd(int a , int b , int c , int d = 0){
    Ed[++cntEd].end = b;
    Ed[cntEd].upEd = head[a];
    Ed[cntEd].f = c;
    Ed[cntEd].c = d;
    head[a] = cntEd;
}

inline bool SPFA(){
    memset(dis , 0x3f , sizeof(dis));
    dis[S] = 0;
    while(!q.empty())
        q.pop();
    q.push(S);
    flo[S] = INF;
    while(!q.empty()){
        int t = q.front();
        q.pop();
        vis[t] = 0;
        for(int i = head[t] ; i ; i = Ed[i].upEd)
            if(Ed[i].f && dis[Ed[i].end] > dis[t] + Ed[i].c){
                dis[Ed[i].end] = dis[t] + Ed[i].c;
                flo[Ed[i].end] = min(Ed[i].f , flo[t]);
                pre[Ed[i].end] = i;
                if(!vis[Ed[i].end]){
                    vis[Ed[i].end] = 1;
                    q.push(Ed[i].end);
                }
            }
    }
    return dis[T] != dis[T + 1];
}

int EK(){
    int ans = 0;
    while(SPFA()){
        int cur = T , sum = 0;
        while(cur != S){
            sum += Ed[pre[cur]].c;
            Ed[pre[cur]].f -= flo[T];
            Ed[pre[cur] ^ 1].f += flo[T];
            cur = Ed[pre[cur] ^ 1].end;
        }
        ans += sum * flo[T];
    }
    return ans;
}

void input(){
    N = read();
    int cnt = N;
    for(int i = 1 ; i <= N ; ++i){
        for(int j = 1 ; j <= i ; ++j)
            read();
        for(int j = i + 1 ; j <= N ; ++j){
            int a = read();
            addEd(S , ++cnt , 1);
            addEd(cnt , S , 0);
            if(a != 1){
                addEd(cnt , j , 1);
                addEd(j , cnt , 0);
            }
            if(a){
                addEd(cnt , i , 1);
                addEd(i , cnt , 0);
            }
        }
    }
    T = ++cnt;
    for(int i = 1 ; i <= N ; ++i)
        for(int j = 1 ; j <= N ; ++j){
            addEd(i , T , 1 , j * (j - 1) / 2 - (j - 1) * (j - 2) / 2);
            addEd(T , i , 0 , -(j * (j - 1) / 2 - (j - 1) * (j - 2) / 2));
        }
}

void work(){
    cout << N * (N - 1) * (N - 2) / 6 - EK() << endl;
    int cnt = N;
    for(int i = 1 ; i <= N ; ++i)
        for(int j = 1 ; j <= N ; ++j)
            if(i != j)
                if(i > j)
                    ans[i][j] = ans[j][i] ^ 1;
                else{
                    ++cnt;
                    for(int k = head[cnt] ; k ; k = Ed[k].upEd)
                        if(!Ed[k].f){
                            ans[i][j] = Ed[k].end == i;
                            break;
                        }
                }
    for(int i = 1 ; i <= N ; ++i){
        for(int j = 1 ; j <= N ; ++j)
            cout << ans[i][j] << ' ';
        cout << endl;
    }
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in" , "r" , stdin);
    //freopen("out" , "w" , stdout);
#endif
    input();
    work();
    return 0;
}