Bzoj3171 [TJOI2013] 循环格

Description

https://www.luogu.com.cn/problem/P3965

Solution

一开始做完全没有思路啊,状压感觉不太行。考虑费用流。

观察到合法的充要条件是所有点都有且仅有一条入边。反证很好证明。

所以直接跑费用流就好了。点数 $2nm$。

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#include<bits/stdc++.h>
#define int long long
#define ull unsigned long long
#define maxn 5005
#define put() putchar('\n')
#define Tp template<typename T>
#define Ts template<typename T,typename... Ar>
using namespace std;
Tp void read(T &x){
int f=1;x=0;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
x*=f;
}
namespace Debug{
Tp void _debug(char* f,T t){cerr<<f<<'='<<t<<endl;}
Ts void _debug(char* f,T x,Ar... y){while(*f!=',') cerr<<*f++;cerr<<'='<<x<<",";_debug(f+1,y...);}
#define gdb(...) _debug((char*)#__VA_ARGS__,__VA_ARGS__)
}using namespace Debug;
#define fi first
#define se second
#define mk make_pair
const int mod=1e9+7;
int power(int x,int y=mod-2) {
int sum=1;
while (y) {
if (y&1) sum=sum*x%mod;
x=x*x%mod;y>>=1;
}
return sum;
}
int head=1,h[maxn];
struct yyy {
int to,z,w,val;
void add(int x,int y,int W,int V) {
to=y;z=h[x];h[x]=head;w=W;val=V;
}
}a[100005];
void ins(int x,int y,int z,int val) {
a[++head].add(x,y,z,val);
a[++head].add(y,x,0,-val);
}
int s,t,n,m;
int ans,anss;
const int inf=1e9;
namespace Dinic {
int vis[maxn],dis[maxn],now[maxn];
queue<int>q;
bool spfa(void) {
memset(dis,0x3f,sizeof(dis));
memcpy(now,h,sizeof(now));
int i,x;
q.push(s);dis[s]=0;
while (!q.empty()) {
x=q.front();q.pop();
for (i=h[x];i;i=a[i].z) if (a[i].w&&dis[a[i].to]>dis[x]+a[i].val) {
dis[a[i].to]=dis[x]+a[i].val;
q.push(a[i].to);
}
}
return dis[t]<=inf;
}
int anss,ans;
int dfs(int x,int in) {
if (x==t||!in) return anss+=in*dis[t],ans+=in,in;
vis[x]=1;
int i,sum=0,res=in;
for (i=now[x];i&&res;i=a[i].z) {
now[x]=i;
if (!vis[a[i].to]&&dis[a[i].to]==dis[x]+a[i].val&&a[i].w) {
sum=dfs(a[i].to,min(res,a[i].w));
a[i].w-=sum,res-=sum,a[i^1].w+=sum;
}
}
vis[x]=0;
return in-res;
}
pair<int,int> dinic(void) {
while (spfa()) dfs(s,inf);
return mk(ans,anss);
}
}
int fx[5]={0,-1,1,0,0};
int fy[5]={0,0,0,-1,1};
int id(int i,int j) {
if (i==0) i=n;
if (i==n+1) i=1;
if (j==0) j=m;
if (j==m+1) j=1;
return (i-1)*m+j;
}
signed main(void){
int i,j,k,x,y,z,val;char ch;
read(n);read(m);
s=0;t=n*m*2+1;
for (i=1;i<=n;i++) {
for (j=1;j<=m;j++) {
cin>>ch;int op=0;
if (ch=='U') op=1;
else if (ch=='D') op=2;
else if (ch=='L') op=3;
else op=4;
ins(s,id(i,j),1,0);
ins(id(i,j)+n*m,t,1,0);
for (k=1;k<=4;k++) ins(id(i,j),id(i+fx[k],j+fy[k])+n*m,1,(k!=op));
}
}
auto tmp=Dinic::dinic();
// printf("%lld %lld\n",tmp.fi,tmp.se);
gdb(tmp.fi);
printf("%lld\n",tmp.se);
return 0;
}