2018-2019 XIX Open Cup, Grand Prix of Korea B

B - Dev, Please Add This!

思路：

对于每一个经过 '*' 的横线和竖线看成一个整体，设他们分别为分量x和分量y

用2-sat考虑这个问题，

如果要经过 '*' ，那么x和y至少要有一个，即连边 !x -> y，!y -> x

如果对于某个分量a不能从通过 'O' 的分量到达它，那么连边 a -> !a

如果对于两个分量a、b，不能同时到达，那么连边a -> !b，b -> !a

然后用2-sat判断可不可行就可以了

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb emplace_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "
";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head
 
const int N = 55, M = 2555;
int n, m, ST, id1[N][N], id2[N][N], cnt, x, y;
pii st[M], ed[M];
bool can[M][M];
char s[N][N];
namespace two_sat {
    int dfn[M*2], low[M*2], cnt, stk[M*2], top, cmp[M*2], tot, n;
    bool vis[M*2];
    vector<int> g[M*2];
    void init(int sz) {
        n = sz;
    }
    void add(int u, int v) {
        g[u].pb(v);
    }
    void tarjan(int u) {
        dfn[u] = low[u] = ++cnt;
        stk[++top] = u;
        vis[u] = true;
        for (int v : g[u]) {
            if(!dfn[v]) tarjan(v), low[u] = min(low[u], low[v]);
            else if(vis[v]) low[u] = min(low[u], dfn[v]);
        }
        if(dfn[u] == low[u]) {
            cmp[u] = ++tot;
            while(stk[top] != u) cmp[stk[top]] = tot, vis[stk[top--]] = false;
            vis[stk[top--]] = false;
        }
    }
    bool ck() {
        for (int i = 1; i <= 2*n; ++i) if(!dfn[i]) tarjan(i);
        for (int i = 1; i <= n; ++i) {
            if(cmp[i] == cmp[i+n]) return false;
        }
        return true;
    }
}
void dfs(int u) {
    can[ST][u] = true;
    int v = id1[st[u].fi][st[u].se]^id2[st[u].fi][st[u].se]^u;
    if(!can[ST][v]) dfs(v);
    v = id1[ed[u].fi][ed[u].se]^id2[ed[u].fi][ed[u].se]^u;
    if(!can[ST][v]) dfs(v);
}
int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%s", s[i]+1);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            if(s[i][j] == 'O') x = i, y = j;
            if(s[i][j] != '#') {
                if(j == 1 || s[i][j-1] == '#') st[++cnt] = {i, j};
                id1[i][j] = cnt;
                if(j == m || s[i][j+1] == '#') ed[cnt] = {i, j};
            }
        }
    }
    for (int j = 1; j <= m; ++j) {
        for (int i = 1; i <= n; ++i) {
            if(s[i][j] != '#') {
                if(i == 1 || s[i-1][j] == '#') st[++cnt] = {i, j};
                id2[i][j] = cnt;
                if(i == n || s[i+1][j] == '#') ed[cnt] = {i, j};
            }
        }
    }
    two_sat::init(cnt);
    for (int i = 1; i <= cnt; ++i) {
        ST = i;
        dfs(ST);
    }
    for (int i = 1; i <= cnt; ++i) if(!can[id1[x][y]][i] && !can[id2[x][y]][i]) two_sat::add(i, i+cnt);
    for (int i = 1; i <= cnt; ++i) {
        for (int j = i+1; j <= cnt; ++j) {
            if(!can[i][j] && !can[j][i]) two_sat::add(i, cnt+j), two_sat::add(j, cnt+i);
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            if(s[i][j] == '*') {
                two_sat::add(id1[i][j]+cnt, id2[i][j]);
                two_sat::add(id2[i][j]+cnt, id1[i][j]);
            }
        }
    }
    if(two_sat::ck()) printf("YES
");
    else printf("NO
");
    return 0;
}