[bzoj4554] [Tjoi2016&Heoi2016]游戏

Posted 2020-09-03 CHADLZX

传送门

网络流的题目做的还真不是很多,这种题目很容易看出是网络流,但就是不怎么会建图.

如果没有硬石头,就是一个经典的二分图匹配问题.

但是有硬石头存在,由于硬石头对前后左右的状态不会发生传递,因此,可以以硬石头为界建立联通块.

然后跑网络流就行了.

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<string>
#include<iomanip>
#include<algorithm>
#include<map>
using namespace std;
#define LL long long
#define FILE "dealing"
#define up(i,j,n) for(int i=j;i<=n;++i)
#define db double
#define ull unsigned long long
#define eps 1e-10
#define pii pair<int,int>
int read(){
	int x=0,f=1,ch=getchar();
	while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
	while(ch>=‘0‘&&ch<=‘9‘){x=(x<<1)+(x<<3)+ch-‘0‘;ch=getchar();}
	return f*x;
}
const int maxn=53,maxm=20000,mod=(int)(1e9+7+0.1),limit=(int)(1e6+1),inf=(int)(1e9);
int n,m,S,T;
char s[53][53];
struct node{
	int y,next,flow,rev;
}e[maxm];
int len,linkk[maxm];
void insert(int x,int y,int flow){
	e[++len].y=y;
	e[len].next=linkk[x];
	linkk[x]=len;
	e[len].flow=flow;
	e[len].rev=len+1;
	e[++len].y=x;
	e[len].next=linkk[y];
	linkk[y]=len;
	e[len].flow=0;
	e[len].rev=len-1;
}
int heng[maxn][maxn],shu[maxn][maxn],cnt1,cnt2;
int q[maxm],head,tail,dis[maxm];
bool makelevel(){
	head=tail=0;
	memset(dis,-1,sizeof(dis));
	q[++tail]=S;dis[S]=0;
	while(++head<=tail){
		int x=q[head];
		for(int i=linkk[x];i;i=e[i].next){
			if(e[i].flow&&dis[e[i].y]==-1){
				dis[e[i].y]=dis[x]+1;
				q[++tail]=e[i].y;
			}
		}
	}
	return dis[T]!=-1;
}
int makeflow(int x,int flow){
	if(x==T)return flow;
	int maxflow=0,d=0;
	for(int i=linkk[x];i&&maxflow!=flow;i=e[i].next){
		if(e[i].flow&&dis[e[i].y]==dis[x]+1)
			if(d=makeflow(e[i].y,min(e[i].flow,flow-maxflow))){
				maxflow+=d;
				e[i].flow-=d;
				e[e[i].rev].flow+=d;
			}
	}
	if(!maxflow)dis[x]=-1;
	return maxflow;
}
int ans=0;
void dinic(){
	int d;
	while(makelevel())
		while(d=makeflow(S,inf))
			ans+=d;
}
int main(){
	freopen(FILE".in","r",stdin);
	freopen(FILE".out","w",stdout);
	n=read();m=read();S=0;
	up(i,1,n)up(j,1,m)scanf(" %c",&s[i][j]);
	for(int i=1;i<=n;i++){
		cnt1++;
		for(int j=1;j<=n;j++){
			if((s[i][j]==‘*‘||s[i][j]==‘x‘)&&s[i][j-1]==‘#‘)cnt1++;
			if(s[i][j]==‘*‘)
				heng[i][j]=cnt1;
		}
	}
	cnt2=cnt1;
	for(int j=1;j<=m;j++){
		cnt2++;
		for(int i=1;i<=n;i++){
			if((s[i][j]==‘*‘||s[i][j]==‘x‘)&&s[i-1][j]==‘#‘)cnt2++;
			if(s[i][j]==‘*‘){
				shu[i][j]=cnt2;
				insert(heng[i][j],shu[i][j],1);
			}
		}
	}
	T=cnt1+cnt2+1;
	up(i,1,cnt1)insert(S,i,1);
	up(i,1,cnt2)insert(i+cnt1,T,1);
	dinic();
	printf("%d\n",ans);
	return 0;
}