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這個博弈類似放骨牌，參見這道題
所以就可以黑白染色之後跑二分圖最大匹配，其中的不必匹配的點就是答案
這些點是什麽呢，\(yy\) 一下發現貌似就是殘余網絡中與 \(s\) 或 \(t\) 在同一個強連通分量的點?

# 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 (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;
}
 

BZOJ1443: [JSOI2009]遊戲Game