[BJOI2019省内集训]完美塔防 题解

没学过2-SAT... 亏爆
考虑把每个炮台当做一个01变量，0横着放，1竖着放，再把图转成约束条件。
具体来说：
如果某一种摆放方式能打到炮台：强制其为false, 即\(addedge(true(x), false(x))\)
某一个空地能被最多两个方向的炮台打到，就可以建立一个or关系。
就做完了（雾
输出方案就是2-SAT的经典问题了，不过代码中没写

#include <bits/stdc++.h>
#define mp make_pair
#define fi first
#define se second
using namespace std;

const int lim = 10000 + 10;
const int N = 100 + 5;
int T, n, m;
vector <int> G[lim];
char c[N][N];
int cov[2][N][N], fix[N][N], scc, instack[lim] , color[lim];
vector <int> v[N][N];
int low[lim], dfn[lim];
pair <int , int> change(int x, int y) {
    swap(x, y);
    if(x != 0) x = -x; if(y != 0) y = -y;
    return make_pair(x, y);
}
bool dfs(int x, int y, int dx, int dy, int id) {
    //cout << x << " " << y << " " << dx << " " << dy << " " << id << endl;
    if(x < 1 || x > n || y < 1 || y > m || c[x][y] == '#') return 1;
    if(c[x][y] == '-' || c[x][y] == '|') return 0;

    cov[id][x][y] = 1;
    if(c[x][y] == '/') {
        pair <int, int> nxt = change(dx, dy);
        return dfs(x + nxt.fi, y + nxt.se, nxt.fi, nxt.se, id);
    }
    else if(c[x][y] == '\\')
        return dfs(x + dy, y + dx, dy, dx, id);
    else if(c[x][y] == '.') return dfs(x + dx, y + dy, dx, dy, id);
}

stack <int> s;
int dfs_clock;
void tarjan(int u) {
    low[u] = dfn[u] = ++dfs_clock;
    s.push(u); instack[u] = 1;
    for(int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if(!dfn[v]) tarjan(v), low[u] = min(low[u], low[v]);
        else if(instack[v]) low[u] = min(low[u], dfn[v]);
    }
    if(low[u] == dfn[u]) {
        ++scc;
        while(1) {
            int t = s.top(); s.pop(); instack[t] = 0;
            color[t] = scc;
            if(t == u) break;
        }
    }
}
void solve() {
    while(!s.empty()) s.pop();
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= 10000; i++) G[i].clear();
    for(int i = 1; i <= n; i++) scanf("%s", (c[i] + 1));
    int cnt = 0;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            fix[i][j] = 0, v[i][j].clear();
    memset(low, 0, sizeof(low));
    memset(dfn, 0, sizeof(dfn));
    memset(color, 0, sizeof(color));
    scc = 0; dfs_clock = 0;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++) {
            if(c[i][j] == '-' || c[i][j] == '|') {
                cnt++;
                memset(cov, 0, sizeof(cov));
                // number : cnt , cnt ^ 1
                //cout << "now dfs:" << i <<" " << j <<endl;
                int id1 = dfs(i, j-1, 0, -1, 0), id2 = dfs(i, j+1, 0, 1, 0);
                int id3 = dfs(i-1, j, -1, 0, 1) , id4 = dfs(i+1, j, 1, 0, 1);
                int num0 = (id1 && id2);
                int num1 = (id3 && id4);
                if(!num0 && !num1) { // 都会攻击到别人
                    puts("IMPOSSIBLE");
                    //cout << i << " " << j << endl;
                    return ;
                }
                    // 2 sat 中强制x为false:  (~x | ~x)
                    // x = true -> x = false
                else if(!num0) {
                    for(int k = 1; k <= n; k++)
                        for(int l = 1; l <= m; l++) {
                            if(cov[1][k][l] == 1 && c[k][l] == '.') v[k][l].push_back(cnt<<1|1);
                        }
                    G[cnt<<1].push_back(cnt<<1|1);
                }
                else if(!num1) {
                    for(int k = 1; k <= n; k++)
                        for(int l = 1; l <= m; l++)
                            if(cov[0][k][l] == 1 && c[k][l] == '.') v[k][l].push_back(cnt<<1);
                    G[cnt<<1|1].push_back(cnt<<1);
                }
                else {
                    for(int k = 1; k <= n; k++) {
                        for(int l = 1; l <= m; l++) if(c[k][l] == '.'){
                            if(cov[0][k][l] && cov[1][k][l]) fix[k][l] = 1;
                            else if(cov[0][k][l]) v[k][l].push_back(cnt<<1);
                            else if(cov[1][k][l]) v[k][l].push_back(cnt<<1|1);
                        }
                    }
                }
            }
        }
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) if(c[i][j] == '.' && fix[i][j] != 1){
            if(v[i][j].size() == 0) {
                puts("IMPOSSIBLE");
                return ;
            }
            else if(v[i][j].size() == 1) {
                G[v[i][j][0]^1].push_back(v[i][j][0]);
            }
            else if(v[i][j].size() == 2) {
                // 经典的u or v
                int uu = v[i][j][0], vv =  v[i][j][1];
                G[uu^1].push_back(vv); G[vv^1].push_back(uu);
            }
        }
    }


    for(int i = 2; i <= (cnt << 1 | 1); i++) if(!dfn[i]) tarjan(i);
    for(int i = 1; i <= (cnt); i++) if(color[i<<1] == color[i<<1|1]) {
        puts("IMPOSSIBLE");
        return ;
    }
    puts("POSSIBLE");
}
int main() {
    scanf("%d", &T);
    while(T--) solve();
    return 0;
}

标签：BJOI2019,return,int,题解,塔防,dfs,low,dx,dy
来源： https://www.cnblogs.com/LiM-817/p/10887160.html