BZOJ1443: [JSOI2009]游戏Game

最新推荐文章于 2020-04-01 20:39:54 发布

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最新推荐文章于 2020-04-01 20:39:54 发布

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分类专栏：网络流博弈论

本文链接：https://blog.youkuaiyun.com/oi_Konnyaku/article/details/86489989

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博弈论

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本文探讨了在类似放骨牌的博弈问题中，如何通过黑白染色和二分图最大匹配算法找到最优策略。特别地，文章详细介绍了如何构建残余网络，并在强连通分量中寻找不必匹配的点作为解答，同时考虑了特殊情况下的单独联通块。

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传送门
这个博弈类似放骨牌，参见这道题
所以就可以黑白染色之后跑二分图最大匹配，其中的不必匹配的点就是答案
这些点是什么呢， $y y$ 一下发现貌似就是残余网络中与 $s$ 或 $t$ 在同一个强连通分量的点?
还要特判一下只有联通块只有一个点的点

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn(10005);
const int maxm(1e6 + 5);

int first[maxn], id[105][105], n, m, cnt, idx, lev[maxn], cur[maxn];
int dfn[maxn], low[maxn], dfx, st[maxn], tp, vis[maxn], bel[maxn], mark[maxn];
queue <int> q;
char mp[105][105];

struct Edge {
	int to, next, w;
} edge[maxm];

inline void Add(int u, int v, int w) {
	edge[cnt] = (Edge){v, first[u], w}, first[u] = cnt++;
	edge[cnt] = (Edge){u, first[v], 0}, first[v] = cnt++;
}

inline int Bfs() {
	int e, u, v;
	memset(lev, 0, sizeof(lev));
	lev[0] = 1, q.push(0);
	while (!q.empty()) {
		u = q.front(), q.pop();
		for (e = first[u]; ~e; e = edge[e].next)
			if (!lev[v = edge[e].to] && edge[e].w) lev[v] = lev[u] + 1, q.push(v);
	}
	return lev[idx + 1];
}

int Dfs(int u, int maxf) {
	if (u == idx + 1) return maxf;
	int ret, &e = cur[u], v, f;
	for (ret = 0; ~e; e = edge[e].next)
		if (lev[v = edge[e].to] == lev[u] + 1 && edge[e].w) {
			f = Dfs(v, min(edge[e].w, maxf - ret));
			ret += f, edge[e].w -= f, edge[e ^ 1].w += f;
			if (ret == maxf) break;
		}
	if (!ret) lev[u] = 0;
	return ret;
}

void Tarjan(int u) {
	low[u] = dfn[u] = ++dfx, st[++tp] = u, vis[u] = 1;
	int e, v;
	for (e = first[u]; ~e; e = edge[e].next)
		if (edge[e].w) {
			if (!dfn[v = edge[e].to]) Tarjan(v), low[u] = min(low[u], low[v]);
			else if (vis[v]) low[u] = min(low[u], dfn[v]);
		}
	if (low[u] == dfn[u]) {
		do {
			v = st[tp--], vis[v] = 0;
			bel[v] = u;
		} while (v ^ u);
	}
}

int main() {
	memset(first, -1, sizeof(first));
	int i, j, e, v, ans;
	scanf("%d%d", &n, &m), ans = 0;
	for (i = 1; i <= n; ++i) {
		scanf(" %s", mp[i] + 1);
		for (j = 1; j <= m; ++j) id[i][j] = ++idx;
	}
	for (i = 1; i <= n; ++i)
		for (j = 1; j <= m; ++j)
			if (mp[i][j] != '#' && (~(i + j) & 1)){
				Add(0, id[i][j], 1);
				if (mp[i - 1][j] == '.') Add(id[i][j], id[i - 1][j], 1);
				if (mp[i + 1][j] == '.') Add(id[i][j], id[i + 1][j], 1);
				if (mp[i][j - 1] == '.') Add(id[i][j], id[i][j - 1], 1);
				if (mp[i][j + 1] == '.') Add(id[i][j], id[i][j + 1], 1);
			}
			else if (mp[i][j] != '#') Add(id[i][j], idx + 1, 1);
	while (Bfs()) memcpy(cur, first, sizeof(cur)), Dfs(0, 1e9);
	for (i = 0; i <= idx + 1; ++i) if (!dfn[i]) Tarjan(i);
	ans = 0;
	for (i = 1; i <= n; ++i)
		for (j = 1; j <= m; ++j)
			if (mp[i][j] == '.')
				mark[id[i][j]] |= (mp[i - 1][j] != '.' && mp[i + 1][j] != '.' && mp[i][j - 1] != '.' && mp[i][j + 1] != '.');
	for (e = first[0]; ~e; e = edge[e].next) if (bel[v = edge[e].to] == bel[0]) mark[v] = 1;
	for (e = first[idx + 1]; ~e; e = edge[e].next) if (bel[v = edge[e].to] == bel[idx + 1]) mark[v] = 1;
	for (i = 1; i <= n; ++i)
		for (j = 1; j <= m; ++j)
			if (mark[id[i][j]]) ++ans;
	ans ? puts("WIN") : puts("LOSE");
	for (i = 1; i <= n; ++i)
		for (j = 1; j <= m; ++j)
			if (mark[id[i][j]]) printf("%d %d\n", i, j);
	return 0;
}