题目大意：给出一个表格，每个表格指向周围四个格子中的一个，问你可以改变一些格子上的指向，问让所有格子都在圈中最小需要改变多少。

思路：所有的格子都在圈中，由于每个格子只能有一个出边，所以就要保证所有格子都有一个入边。建立费用流的模型，所有点向汇点连流量1费用0的边，表示要接受一个入边。S向所有点连一条流量1费用0的边，表示一条出边。一个格子向周围四个格子连边，流量1，如果方向与当前方向相符，那么费用0，否则费用1。之后跑费用流就是答案了。

CODE：

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 10010
#define INF 0x3f3f3f3f
#define S 0
#define T (MAX - 1)
using namespace std;
const int dx[] = {0,0,0,-1,1};
const int dy[] = {0,-1,1,0,0};
int G[256];

struct MinCostMaxFlow{
	int head[MAX],total;
	int next[MAX],aim[MAX],cost[MAX],flow[MAX];

	int f[MAX],from[MAX],p[MAX];
	bool v[MAX];

	MinCostMaxFlow() {
		total = 1;
	}
	void Add(int x,int y,int f,int c) {
		next[++total] = head[x];
		aim[total] = y;
		flow[total] = f;
		cost[total] = c;
		head[x] = total;
	}
	void Insert(int x,int y,int f,int c) {
		Add(x,y,f,c);
		Add(y,x,0,-c);
	}
	bool SPFA() {
		static queue<int> q;
		while(!q.empty())	q.pop();
		memset(f,0x3f,sizeof(f));
		memset(v,false,sizeof(v));
		f[S] = 0;
		q.push(S);
		while(!q.empty()) {
			int x = q.front(); q.pop();
			v[x] =  false;
			for(int i = head[x]; i; i = next[i])
				if(flow[i] && f[aim[i]] > f[x] + cost[i]) {
					f[aim[i]] = f[x] + cost[i];
					if(!v[aim[i]])
						v[aim[i]] = true,q.push(aim[i]);
					from[aim[i]] = x;
					p[aim[i]] = i;
				}
		}
		return f[T] != INF;
	}
	int EdmondsKarp() {
		int re = 0;
		while(SPFA()) {
			int max_flow = INF;
			for(int i = T; i != S; i = from[i])
				max_flow = min(max_flow,flow[p[i]]);
			for(int i = T; i != S; i = from[i]) {
				flow[p[i]] -= max_flow;
				flow[p[i]^1] += max_flow;
			}
			re += max_flow * f[T];
		}
		return re;
	}
}solver;

int m,n;
int num[20][20],cnt;
char s[20][20];

int main()
{
	G['L'] = 1,G['R'] = 2,G['U'] = 3,G['D'] = 4;
	cin >> m >> n;
	for(int i = 1; i <= m; ++i) {
		scanf("%s",s[i] + 1);
		for(int j = 1; j <= n; ++j) {
			num[i][j] = ++cnt;
			solver.Insert(S,cnt << 1,1,0);
			solver.Insert(cnt << 1|1,T,1,0);
		}
	}
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j)
			for(int k = 1; k <= 4; ++k) {
				int fx = i + dx[k],fy = j + dy[k];
				if(!fx)	fx = m;
				if(!fy)	fy = n;
				if(fx > m)	fx = 1;
				if(fy > n)	fy = 1;
				if(k == G[s[i][j]])
					solver.Insert(num[i][j] << 1,num[fx][fy] << 1|1,1,0);
				else
					solver.Insert(num[i][j] << 1,num[fx][fy] << 1|1,1,1);
			}
	cout << solver.EdmondsKarp() << endl;
	return 0;
}

BZOJ 3171 TJOI 2013 循环格 费用流