BZOJ2597 WC2007剪刀石頭布（費用流）

　　考慮使非剪刀石頭布情況盡量少。設第i個人贏了x_i場，那么以i作為贏家的非剪刀石頭布情況就為x_i(x_i-1)/2種。那么使Σx_i(x_i-1)/2盡量小即可。

　　考慮網絡流。將比賽建成一排點，人建成一排點，每場未確定比賽向比賽雙方連邊，確定比賽向贏者連邊，這樣就是一種合法的比賽方案了。

　　在此基礎上控制代價最小。由于每多贏一場非剪刀石頭布情況的增量就更大，將邊拆開費用設為增量即可。

#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[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<<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_JUDGEfreopen("bzoj2597.in","r",stdin);freopen("bzoj2597.out","w",stdout);const char LL[]="%I64d\n";
#elseconst char LL[]="%lld\n";
#endifcnt=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;
}

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