P5038 [SCOI2012]奇怪的游戏

题目链接

题意分析

首先我们需要求的是统一以后的值\(x\)

并且一般的棋盘操作我们都需要黑白染色

那么对于棋盘格子是偶数的情况的话

答案是存在单调性的

因为如果统一之后 两两搭配还是可以再加一个的

如果棋盘格子是奇数的话

那么黑格子数量为\(num1\) 权值和为\(sum1\)

白格子数量为\(num2\) 权值和为\(sum2\)

那么

\[num1*x-sum1=num2*x-sum2
\]

然后我们可以使用最大流检验合法性

用源点向黑色的点连边权为\(x-num[i][j]\)的边

白色的点向汇点连边权为\(x-num[i][j]\)的边

然后黑色的点向白色的点连边权为\(inf\)的边

然后检查能否跑满流即可

至于这么做的意义 应该很好理解

也就是相邻两个点之间的匹配问题

CODE:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<string>
#include<queue>
#include<map>
#include<stack>
#include<list>
#include<set>
#include<deque>
#include<vector>
#include<ctime>
#define ll long long
#define inf 1e18
#define N 48
#define IL inline
#define M 1008611
#define D double
#define ull unsigned long long
#define R register
using namespace std;
template<typename T>IL void read(T &_)
{
    T __=0,___=1;char ____=getchar();
    while(!isdigit(____)) {if(____=='-') ___=0;____=getchar();}
    while(isdigit(____)) {__=(__<<1)+(__<<3)+____-'0';____=getchar();}
    _=___ ? __:-__;
}
/*-------------OI使我快乐-------------*/
ll opt,n,m,tot=1,S,T,maxn;
ll sum1,sum2,num1,num2,all;
ll hei[N][N],bel[N][N];
ll to[M],nex[M],head[N*N],fro[M];ll w[M];
ll dep[N*N],cur[N*N];queue<ll> Q;
ll tx[6]={0,0,0,1,-1},ty[6]={0,1,-1,0,0};
IL void add(ll x,ll y,ll z)
{to[++tot]=y;fro[tot]=x;nex[tot]=head[x];head[x]=tot;w[tot]=z;
swap(x,y);to[++tot]=y;fro[tot]=x;nex[tot]=head[x];head[x]=tot;w[tot]=0;}
IL bool safe(ll x,ll y){return x>=1&&x<=n&&y>=1&&y<=m;}
IL ll id(ll x,ll y){return (x-1)*m+y;}
IL bool bfs()
{
	for(R ll i=1;i<=T;++i) dep[i]=0;
	Q.push(S);dep[S]=1;
	for(;!Q.empty();)
	{
		ll u=Q.front();Q.pop();
		for(R ll i=head[u];i;i=nex[i])
		{
			ll v=to[i];
			if(w[i]>0&&dep[v]==0)
			{
				dep[v]=dep[u]+1;Q.push(v);
			}
		}
	}
	return dep[T]!=0;
}
IL ll dfs(ll now,ll res)
{
	if(now==T||!res) return res;
	for(R ll &i=cur[now];i;i=nex[i])
	{
		ll v=to[i];
		if(w[i]>0&&dep[v]==dep[now]+1)
		{
			ll have=dfs(v,min(w[i],res));
			if(have>0)
			{
				w[i]-=have;w[i^1]+=have;
				return have;
			}
		}
	}
	return 0;
}
IL void Dinic()
{
	while(bfs())
	{
//		printf("now nowno\n");
		for(R ll i=1;i<=T;++i) cur[i]=head[i];
		ll d=dfs(S,inf);
//		printf("now now now %lld\n",d);
		while(d) all+=d,d=dfs(S,inf);
	}
}
IL bool check(ll now)
{
	tot=1;all=0;ll tmp=0;
	memset(head,0,sizeof head);
	for(R ll i=1;i<=n;++i)
	 for(R ll j=1;j<=m;++j)
	  if((i+j)&1) add(S,id(i,j),now-hei[i][j]),tmp+=(now-hei[i][j]);
	  else add(id(i,j),T,now-hei[i][j]);
	for(R ll i=1;i<=n;++i)
	 for(R ll j=1;j<=m;++j)
	  if((i+j)&1)
	   for(R ll k=1;k<=4;++k)
	   {

	   	ll nowx=i+tx[k],nowy=j+ty[k];
	   	if(safe(nowx,nowy)) add(id(i,j),id(nowx,nowy),inf);
	   }
//	printf("now is %lld but need is %lld\n",all,tmp);
	Dinic();
//	printf("now is %lld but need is %lld\n",all,tmp);
	return (all==tmp);
}
IL void solve_cdy()
{
	ll le=maxn,ri=1e15,ans=-1;
	while(le<=ri)
	{
//		printf("%lld %lld\n",le,ri);
		ll mid=(le+ri)>>1;
		if(check(mid)) ans=mid,ri=mid-1;
		else le=mid+1;
	}
//	printf("%d\n",ans);
	if(ans==-1) puts("-1");
	else printf("%lld\n",(ans*num1)-sum1);
}
IL void solve_wzy()
{
	if((sum1-sum2)%(num1-num2)==0)
	{
		ll tox=(sum1-sum2)/(num1-num2);
//		printf("tox si %d\n",tox);
		if(tox<1ll*maxn) {puts("-1");return;}
		if(check(tox)) printf("%lld\n",(tox*num1)-sum1);
		else puts("-1");
	}
	else puts("-1");
}
int main()
{
//	freopen("222.in","r",stdin);
//	freopen("222.out","w",stdout);
	read(opt);
	while(opt--)
	{
		maxn=0;sum1=sum2=num1=num2=0;
		read(n);read(m);S=n*m+1;T=n*m+2;
		for(R ll i=1;i<=n;++i)
		 for(R ll j=1;j<=m;++j)
		  read(hei[i][j]),maxn=max(maxn,hei[i][j]);
		for(R ll i=1;i<=n;++i)
		 for(R ll j=1;j<=m;++j)
		 {
		 	bel[i][j]=((i+j)&1);
		 	if(bel[i][j]) num1++,sum1+=hei[i][j];
		 	else num2++,sum2+=hei[i][j];
		 }
//		printf("%lld %lld %lld %lld\n",num1,sum1,num2,sum2);
		if(num1==num2) solve_cdy();
		else solve_wzy();
	}
//	fclose(stdin);
//	fclose(stdout);
    return 0;
}

HEOI 2019 RP++