JZOJ 3281. 【GDOI2013】字母连接

\(\text{Solution}\)

一眼不会，限制有点多。。。
那就网络流
发下确实是很简单的建图
枚举起点集合
拆点后就很好满足限制了

\(\text{Code}\)

#include <cstdio>
#include <iostream>
#include <cstring>
#define RE register
#define IN inline
using namespace std;

const int N = 1005, INF = 2147483647;
int n, m, S, T, num, p1, p2, tot, ans;
int dis[N], h[N], vis[N], pre[N], edge[N], flow[N], a[25][25], d[N], buc[N], Q[N];
int fx[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
char str[20];
struct node{int to, nxt, w, f;}e[N * 10];
IN void add(int u, int v, int w, int f)
{
	e[++tot] = node{v, h[u], w, f}, h[u] = tot;
	e[++tot] = node{u, h[v], 0, -f}, h[v] = tot;
}
IN int Isin(int x, int y){return (x > 0 && x <= n && y > 0 && y <= m && a[x][y] != 101);}
IN int getid(int x, int y){return (x - 1) * m + y;}

IN int spfa()
{
	for(RE int i = S; i <= T; i++) vis[i] = 0, dis[i] = flow[i] = INF;
	int head = 0, tail = 1;
	d[1] = S, vis[S] = 1, dis[S] = 0, pre[T] = -1;
	while (head < tail)
	{
		int now = d[++head];
		vis[now] = 0;
		for(RE int i = h[now]; i; i = e[i].nxt)
		if (dis[e[i].to] > dis[now] + e[i].f && e[i].w)
		{
			dis[e[i].to] = dis[now] + e[i].f, flow[e[i].to] = min(flow[now], e[i].w), pre[e[i].to] = now, edge[e[i].to] = i;
			if (!vis[e[i].to]) vis[e[i].to] = 1, d[++tail] = e[i].to;
		}
	}
	return pre[T] != -1;
}
IN int MCMF()
{
	int now, Maxflow = 0, Mincost = 0;
	while (spfa()) 
	{
		Maxflow += flow[T], Mincost += dis[T] * flow[T], now = T;
		while (now != S) e[edge[now]].w -= flow[T], e[edge[now] ^ 1].w += flow[T], now = pre[now];
	}
	if (Maxflow != num / 2) return INF;
	return Mincost;
}

IN void solve()
{
	memset(h, 0, sizeof h), tot = 1, T = n * m * 2 + 1;
	for(RE int i = 1; i <= n; i++)
		for(RE int j = 1; j <= m; j++)
		if (a[i][j] != 101)
		{
			if (a[i][j] == 100) add(getid(i, j), getid(i, j) + n * m, 1, 1);
			else add(getid(i, j), getid(i, j) + n * m, 1, 0);
			for(RE int k = 0; k < 4; k++)
			{
				int x = i + fx[k][0], y = j + fx[k][1];
				if (Isin(x, y)) add(getid(i, j) + n * m, getid(x, y), 1, 0);
			}
		}
	for(RE int i = 1; i <= num; i++)
		if (buc[Q[i]]) add(S, Q[i], 1, 0);
		else add(Q[i] + n * m, T, 1, 0);
	ans = min(ans, MCMF());
}

void dfs(int x, int s)
{
	if (s > num / 2) return;
	if (x > num)
	{
		if (s == num / 2 && ((!p1 && !p2) || (buc[p1] && buc[p2]))) solve();
		return;
	}
	buc[Q[x]] = 1, dfs(x + 1, s + 1), buc[Q[x]] = 0, dfs(x + 1, s);
}

int main() 
{
	int q; scanf("%d", &q);
	for(; q; --q)
	{
		memset(buc, 0, sizeof buc), num = p1 = p2 = 0, ans = INF, scanf("%d%d", &n, &m);
		for(RE int i = 1; i <= n; i++)
		{
			scanf("%s", str + 1);
			for(RE int j = 1, id; j <= m; j++)
				if (str[j] == '.') a[i][j] = 100;
				else if (str[j] == '#') a[i][j] = 101;
				else
				{
					a[i][j] = str[j] - 'A' + 1, id = getid(i, j), Q[++num] = id;
					if (buc[a[i][j]]) p1 = buc[a[i][j]], p2 = id;
					buc[a[i][j]] = id;
				}
		}
		memset(buc, 0, sizeof buc), dfs(1, 0), printf("%d\n", (ans == INF ? -1 : ans));
	}
}