BZOJ2597 WC2007剪刀石头布（费用流）

　　考虑使非剪刀石头布情况尽量少。设第i个人赢了x场，那么以i作为赢家的非剪刀石头布情况就为x(x-1)/2种。那么使Σx(x-1)/2尽量小即可。

　　考虑网络流。将比赛建成一排点，人建成一排点，每场未确定比赛向比赛双方连边，确定比赛向赢者连边，这样就是一种合法的比赛方案了。

　　在此基础上控制代价最小。由于每多赢一场非剪刀石头布情况的增量就更大，将边拆开费用设为增量即可。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

using namespace std;

int read()

{

    int x=,f=;char c=getchar();

    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}

    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();

    return x*f;

}

#define N 110

#define S 0

#define T 10101

int n,p[N*N],l[N*N],r[N*N],cnt,t=-,ans,a[N][N];

int d[N*N],q[N*N],pre[N*N];

bool flag[N*N];

struct data{int to,nxt,cap,flow,cost;

}edge[N*N<<];

void addedge(int x,int y,int z,int cost)

{

    t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].cap=z,edge[t].flow=,edge[t].cost=cost,p[x]=t;

    t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=,edge[t].flow=,edge[t].cost=-cost,p[y]=t;

}

int inc(int &x){x++;if (x>T) x-=T;return x;}

bool spfa()

{

    memset(d,,sizeof(d));d[S]=;

    memset(flag,,sizeof(flag));

    int head=,tail=;q[]=S;

    do

    {

        int x=q[inc(head)];flag[x]=;

        for (int i=p[x];~i;i=edge[i].nxt)

        if (d[x]+edge[i].cost<d[edge[i].to]&&edge[i].flow<edge[i].cap)

        {

            d[edge[i].to]=d[x]+edge[i].cost;

            pre[edge[i].to]=i;

            if (!flag[edge[i].to]) q[inc(tail)]=edge[i].to,flag[edge[i].to]=;

        }

    }while (head!=tail);

    return d[T]<=;

}

void ekspfa()

{

    while (spfa())

    {

        int v=;

        for (int i=T;i!=S;i=edge[pre[i]^].to)

        if (edge[pre[i]].flow==edge[pre[i]].cap) {v=;break;}

        if (v)

        for (int i=T;i!=S;i=edge[pre[i]^].to)

        ans-=edge[pre[i]].cost,edge[pre[i]].flow++,edge[pre[i]^].flow--;

    }

}

int main()

{

#ifndef ONLINE_JUDGE

    freopen("bzoj2597.in","r",stdin);

    freopen("bzoj2597.out","w",stdout);

    const char LL[]="%I64d\n";

#else

    const char LL[]="%lld\n";

#endif

    cnt=n=read();

    memset(p,,sizeof(p));

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

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

        {

            int x=read();

            if (i<j)

            {

                cnt++;l[cnt]=i,r[cnt]=j;

                addedge(S,cnt,,);

                if (x==) addedge(cnt,j,,);

                else if (x==) addedge(cnt,i,,);

                else addedge(cnt,i,,),addedge(cnt,j,,);

            }

        }

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

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

        addedge(i,T,,j-);

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

    ekspfa();

    cout<<ans<<endl;

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

        for (int j=p[i];~j;j=edge[j].nxt)

        if (edge[j].flow>)

            if (edge[j].to==l[i]) a[l[i]][r[i]]=;else a[r[i]][l[i]]=;

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

    {

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

        printf("%d ",a[i][j]);

        printf("\n");

    }

    return ;

}