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  • BZOJ1443: [JSOI2009]游戏Game

    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,(yy) 一下发现貌似就是残余网络中与 (s)(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
    ", i, j);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/10271014.html
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