bzoj:3171: [Tjoi2013]循环格

原题链接：http://www.lydsy.com/JudgeOnline/problem.php?id=3171

感觉构图的方法还是很巧妙的。每个格子拆出点入点，分别连到源汇，然后相邻的格子再连一波，其实也是利用了每个格子出度入度均为1。

拉个模版下来都能套错……

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 21000
#define MAXM 12000
using namespace std;

struct na{
    int y,z,f,ne;
}b[MAXM];
int n,m,l[MAXN],r[MAXN],num=1,p,ch,ans=0,S,T,k,dis[MAXN],an=0,mi[MAXN],ro[MAXN],qi[MAXN],a,d,fa,fb,f,v;
bool bo[MAXN];
const int INF=1e9;
queue <int> q;
inline int min(int x,int y){return x>y?y:x;}
inline int read(){
    p=0;ch=getchar();
    while (ch<'0'||ch>'9') ch=getchar();
    while (ch>='0'&&ch<='9') p=p*10+ch-48, ch=getchar();
    return p;
}
char s[20];
inline void spfa(){
    register int i;
    q.push(S);
    bo[S]=1;
    for (i=0;i<=T;i++) dis[i]=INF,mi[i]=0;
    mi[S]=INF;dis[S]=0;
    while(!q.empty()){
        int k=q.front();q.pop();bo[k]=0;
        if (k==T) continue;
        for (i=l[k];i;i=b[i].ne)
        if (b[i].z>0&&dis[b[i].y]>b[i].f+dis[k]){
            dis[b[i].y]=b[i].f+dis[k];
            mi[b[i].y]=min(mi[k],b[i].z);
            ro[b[i].y]=i;
            qi[b[i].y]=k;
            if (!bo[b[i].y]){
                bo[b[i].y]=1;
                q.push(b[i].y);
            }
        }
    }
}
inline int add(int x,int y,int z,int f){
    num++;
    if (l[x]==0) l[x]=num;else b[r[x]].ne=num;
    b[num].y=y;b[num].z=z;b[num].f=f;r[x]=num;
}
inline void in(int x,int y,int z,int f){/*printf("%d %d %d %d
",x,y,z,f);*/add(x,y,z,f),add(y,x,0,-f);}
int main(){
    //freopen("a.in","r",stdin);
    //freopen("a.out","w",stdout);
    register int i,j;
    n=read();m=read();S=n*m*2;T=n*m*2+1;
    for (i=0;i<n;i++)
    for (scanf("%s",s),j=0;j<m;j++){
        in(S,i*m+j,1,0);
        in(i*m+j+n*m,T,1,0);
        in(i*m+j,(i+1)%n*m+j+n*m,1,s[j]!='D');
        in(i*m+j,(i+n-1)%n*m+j+n*m,1,s[j]!='U');
        
        in(i*m+j,i*m+(j+1)%m+n*m,1,s[j]!='R');
        in(i*m+j,i*m+(j+m-1)%m+n*m,1,s[j]!='L');
    }
    
    for(;;){
        spfa();
        if (dis[T]==INF) break;
        an+=mi[T]*dis[T];
        for (i=T;i!=S;i=qi[i]) b[ro[i]].z-=mi[T],b[ro[i]^1].z+=mi[T];
    }
    printf("%d
",an);
}