bzoj 3171: [Tjoi2013]循环格 最小费用最大流

题目大意:

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

题解:

首先我们很容易发现一个结论:

出现完美循环当且仅当所有点的出入度均为1

所以利用这个性质,我们将每个点拆成两个点

S连向出点,入点连向T,然后出点向上下左右的入点连边

跑最小费用最大流即可.

#include <cstdio>
#include <cstring>
#include <cassert>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(int &x){
    x=0;char ch;bool flag = false;
    while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
    while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
inline int cat_max(const int &a,const int &b){return a>b ? a:b;}
inline int cat_min(const int &a,const int &b){return a<b ? a:b;}
const int maxnode = 512;
const int maxedge = 4096;
struct Edge{
    int to,next,cap,cost;
}G[maxedge<<1];
int head[maxnode],cnt=1;
void add(int u,int v,int c,int d){
    G[++cnt].to = v;
    G[cnt].next = head[u];
    head[u] = cnt;
    G[cnt].cap = c;
    G[cnt].cost = d;
}
inline void insert(int u,int v,int c,int d){
    add(u,v,c,d);add(v,u,0,-d);
}
#define v G[i].to
const int lim = maxnode<<1;
int dis[maxnode],p[maxnode],q[lim + 10],flow[maxnode];
int l,r,S,T,ans;bool inq[maxnode];
const int inf = 0x3f3f3f3f;
bool spfa(){
    memset(dis,0x3f,sizeof dis);
    dis[S] = 0;flow[S] = inf;
    inq[S] = true;l = 0;r = -1;
    q[++r] = S;
    while(l <= r){
        int u = q[l % lim];++l;
        for(int i = head[u];i;i=G[i].next){
            if(dis[v] > dis[u] + G[i].cost && G[i].cap){
                dis[v] = dis[u] + G[i].cost;
                flow[v] = min(flow[u],G[i].cap);
                p[v] = i;
                if(!inq[v]){
                    q[++r % lim] = v;
                    inq[v] = true;
                }
            }
        }inq[u] = false;
    }if(dis[T] == dis[0]) return false;
    ans += dis[T]*flow[T];
    for(int u = T;u != S;u = G[p[u]^1].to){
        G[p[u]].cap -= flow[T],G[p[u]^1].cap += flow[T];
    }
    return true;
}
#undef v
#define f(x,y) ((x-1)*m + y)
int dx[] = {0,0,1,-1};
int dy[] = {1,-1,0,0};
inline int id(const char &ch){
    if(ch == 'U') return 3;
    if(ch == 'D') return 2;
    if(ch == 'L') return 1;
    if(ch == 'R') return 0;
    return -1;
}
int main(){
    int n,m;read(n);read(m);
    S = maxnode - 5;T = S+1;
    char ch;
    for(int i=1,x;i<=n;++i){
        for(int j=1;j<=m;++j){
            insert(f(i,j)<<1,T,1,0);
            insert(S,f(i,j)<<1|1,1,0);
            while(ch=getchar(),ch<'!');
            x = id(ch);if(x == -1) assert(0);
            for(int k=0;k<4;++k){
                int nx = i + dx[k];if(nx == n+1) nx = 1;if(nx == 0) nx = n;
                int ny = j + dy[k];if(ny == m+1) ny = 1;if(ny == 0) ny = m;
                insert(f(i,j)<<1|1,f(nx,ny)<<1,1,x != k);
            }
        }
    }
    while(spfa());
    printf("%d\n",ans);
    getchar();getchar();
    return 0;
}