[bzoj4554] [Tjoi2016&Heoi2016]游戏

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;
}