BZOJ 3171 循环格

重点：如果满流一定存在许多个欧拉回路。于是费用流。

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#define maxv 500
#define maxe 1000050
#define maxn 20
#define inf 2000000000
using namespace std;
int n,m,map[maxn][maxn],dis[maxv],nume=1,g[maxv],s,t,pree[maxv],prev[maxv];
int dx[]={0,0,0,-1,1},dy[]={0,-1,1,0,0};
char pp[maxn];
bool vis[maxv];
queue <int> q;
struct edge
{
    int v,f,c,nxt;
}e[maxe];
void get(int x,int y)
{
    if (pp[y-1]=='L') map[x][y]=1;
    else if (pp[y-1]=='R') map[x][y]=2;
    else if (pp[y-1]=='U') map[x][y]=3;
    else map[x][y]=4;
}
void addedge(int u,int v,int f,int c)
{
    e[++nume].v=v;e[nume].f=f;e[nume].c=c;e[nume].nxt=g[u];g[u]=nume;
    e[++nume].v=u;e[nume].f=0;e[nume].c=-c;e[nume].nxt=g[v];g[v]=nume;
}
int p(int x,int y)
{
    return (x-1)*m+y;
}
void build()
{
    s=0;t=2*n*m+1;
    for (int i=1;i<=n*m;i++)
    {
        addedge(s,i,1,0);
        addedge(i+n*m,t,1,0);
    }
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
            for (int k=1;k<=4;k++)
            {
                int tx=(i+dx[k]+n)%n,ty=(j+dy[k]+m)%m;
                if (!tx) tx=n;if (!ty) ty=m;
                if ((tx==i) && (ty==j)) continue;
                if (map[i][j]==k) addedge(p(i,j),n*m+p(tx,ty),1,0);
                else addedge(p(i,j),n*m+p(tx,ty),1,1);
            }
}
bool spfa()
{
    for (int i=s;i<=t;i++) {pree[i]=prev[i]=-1;dis[i]=inf;vis[i]=false;}
    q.push(s);vis[s]=true;dis[s]=0;
    while (!q.empty())
    {
        int head=q.front();q.pop();
        for (int i=g[head];i;i=e[i].nxt)
        {
            int v=e[i].v;
            if ((e[i].f) && (dis[v]>dis[head]+e[i].c))
            {
                dis[v]=dis[head]+e[i].c;
                pree[v]=i;prev[v]=head;
                if (!vis[v]) {q.push(v);vis[v]=true;}
            }
        }
        vis[head]=false;
    }
    if (dis[t]==inf) return false;
    return true;
}
int dinic()
{
    int u=t,low=inf;
    while (u!=s)
    {
        low=min(low,e[pree[u]].f);
        u=prev[u];
    }
    u=t;
    while (u!=s)
    {
        e[pree[u]].f-=low;e[pree[u]^1].f+=low;
        u=prev[u];
    }
    return dis[t]*low;
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++)
    {
        scanf("%s",pp);
        for (int j=0;j<m;j++) get(i,j+1);
    }
    build();
    int min_cost=0;
    while (spfa()) min_cost+=dinic();
    printf("%d\n",min_cost);
    return 0;
}