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xi场，那么以i作为赢家的非剪刀石头布情况就为xi(xi-1)/2种。那么使Σxi(xi-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=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),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=-1,ans,a

;
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<<4];
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=0,edge[t].cost=cost,p[x]=t;
t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=0,edge[t].flow=0,edge[t].cost=-cost,p[y]=t;
}
int inc(int &x){x++;if (x>T) x-=T;return x;}
bool spfa()
{
memset(d,42,sizeof(d));d[S]=0;
memset(flag,0,sizeof(flag));
int head=0,tail=1;q[1]=S;
do
{
int x=q[inc(head)];flag[x]=0;
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]=1;
}
}while (head!=tail);
return d[T]<=10000000;
}
void ekspfa()
{
while (spfa())
{
int v=1;
for (int i=T;i!=S;i=edge[pre[i]^1].to)
if (edge[pre[i]].flow==edge[pre[i]].cap) {v=0;break;}
if (v)
for (int i=T;i!=S;i=edge[pre[i]^1].to)
ans-=edge[pre[i]].cost,edge[pre[i]].flow++,edge[pre[i]^1].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,255,sizeof(p));
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
{
int x=read();
if (i<j)
{
cnt++;l[cnt]=i,r[cnt]=j;
addedge(S,cnt,1,0);
if (x==0) addedge(cnt,j,1,0);
else if (x==1) addedge(cnt,i,1,0);
else addedge(cnt,i,1,0),addedge(cnt,j,1,0);
}
}
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
addedge(i,T,1,j-1);
ans=n*(n-1)*(n-2)/6;
ekspfa();
cout<<ans<<endl;
for (int i=n+1;i<=cnt;i++)
for (int j=p[i];~j;j=edge[j].nxt)
if (edge[j].flow>0)
if (edge[j].to==l[i]) a[l[i]][r[i]]=1;else a[r[i]][l[i]]=1;
for (int i=1;i<=n;i++)
{
for (int j=1;j<=n;j++)
printf("%d ",a[i][j]);
printf("\n");
}
return 0;
}