JZOJ 1083. 【GDOI2006】拯救亚特兰蒂斯

\(\text{Solution}\)

自己的网络流技术太拉了

连这样的题都做不出来

对于一个怪物，剑术和法术两样东西有一样就可以了

不难想到二分图中最小点覆盖，一条边只有两个端点之一被选就被覆盖了

最小点覆盖等于最大匹配数

\(\text{Code}\)

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

const int N = 11005, INF = 1e9;
int p, n, m, S, T, tot, h[N], dep[N], cur[N], Q[N], bz[N];
vector<int> g[N][2];
map<string, int> mp;
char s[20];
struct edge{int to, nxt, w;}e[N * 10];
IN void add(int x, int y, int w)
{
	e[++tot] = edge{y, h[x], w}, h[x] = tot;
	e[++tot] = edge{x, h[y], 0}, h[y] = tot;
}

IN int bfs()
{
	for(RE int i = S; i <= T; i++) cur[i] = h[i], dep[i] = 0;
	int head = 0, tail = 1; Q[1] = S, dep[S] = 1;
	while (head < tail)
	{
		int now = Q[++head];
		for(RE int i = h[now]; i; i = e[i].nxt)
		{
			int v = e[i].to;
			if (dep[v] || !e[i].w) continue;
			dep[v] = dep[now] + 1, Q[++tail] = v;
		}
	}
	return dep[T];
}
int dfs(int x, int lim)
{
	if (x == T || lim <= 0) return lim;
	int flow = 0;
	for(RE int i = cur[x], v, f; i; i = e[i].nxt)
	{
		cur[x] = i, v = e[i].to;
		if (dep[v] != dep[x] + 1 || !e[i].w) continue;
		f = dfs(v, min(lim, e[i].w));
		if (f <= 0) continue;
		e[i].w -= f, e[i ^ 1].w += f, lim -= f, flow += f;
		if (lim <= 0) break;
	}
	return flow;
}
IN void dinic()
{
	int flow = 0;
	while (bfs()) flow += dfs(S, INF);
	printf("%d\n", flow);
}

int main()
{
	scanf("%d%d%d", &p, &n, &m);
	while (1)
	{
		if (n == -1 && m == -1 && p == -1) break;
		T = 0, tot = 1, memset(h, 0, sizeof h); int cnt = 0;
		for(RE int i = 1; i <= p; i++) g[i][0].clear(), g[i][1].clear(), bz[i] = 0;

		for(RE int i = 1; i <= p; i++) scanf("%s", s), mp[s] = ++cnt;
		for(RE int i = 1, num, z; i <= n; i++)
		{
			scanf("%s%d", s, &num);
			for(RE int j = 1; j <= num; j++) scanf("%s", s), g[z = mp[s]][0].push_back(i), bz[z] = 1;
		}
		for(RE int i = 1, num, z; i <= m; i++)
		{
			scanf("%s%d", s, &num), ++T;
			for(RE int j = 1; j <= num; j++) scanf("%s", s), g[z = mp[s]][1].push_back(i + n), bz[z] = 1;
		}
		int flag = 0;
		for(RE int i = 1; i <= p; i++) if (!bz[i]){flag = 1; printf("Poor Atlantis!\n"); break;};
		if (!flag)
		{
			T = n + m + 1;
			for(RE int i = 1; i <= n; i++) add(S, i, 1);
			for(RE int i = 1; i <= m; i++) add(i + n, T, 1);
			for(RE int i = 1; i <= p; i++)
				for(RE int j = 0; j < g[i][0].size(); j++)
					for(RE int k = 0; k < g[i][1].size(); k++) add(g[i][0][j], g[i][1][j], 1);
			dinic();
		}
		scanf("%d%d%d", &p, &n, &m);
	}
}