其实写一个spfa就行 遍历所有的草的点 组合两个
求最大时间的最小
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <list> #include <cmath> #include <bitset> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define rb(a) scanf("%lf", &a) #define rf(a) scanf("%f", &a) #define pd(a) printf("%d ", a) #define plld(a) printf("%lld ", a) #define pc(a) printf("%c ", a) #define ps(a) printf("%s ", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 110000, INF = 0x7fffffff; int n, m; string str[15]; int dis[4][2] = {{1, 0},{-1, 0},{0, 1},{0, -1}}; int vis[15][15], vis1[15][15], d[15][15]; void dfs(int x, int y, int cnt) { for(int i = 0; i < 4; i++) { int nx = x + dis[i][0]; int ny = y + dis[i][1]; if(nx < 0 || nx >= n || ny < 0 || ny >=m || str[nx][ny] == '.' || vis[nx][ny]) continue; vis[nx][ny] = cnt; dfs(nx, ny, cnt); } } struct node { int x, y; node(int x, int y) : x(x), y(y) {} }; int spfa(int x, int y) { for(int i = 0; i <= n; i++) for(int j = 0; j <= m; j++) d[i][j] = INF; vector<node> f; mem(vis1, 0); queue<node> Q; Q.push(node(x, y)); vis1[x][y] = 1; d[x][y] = 0; while(!Q.empty()) { node u = Q.front(); Q.pop(); f.push_back(u); vis1[u.x][u.y] = 0; for(int i = 0; i < 4; i++) { node t = node(0, 0); t.x = u.x + dis[i][0]; t.y = u.y + dis[i][1]; if(t.x < 0 || t.x >= n || t.y < 0 || t.y >= m || str[t.x][t.y] == '.') continue; if(d[t.x][t.y] > d[u.x][u.y] + 1) { d[t.x][t.y] = d[u.x][u.y] + 1; if(!vis1[t.x][t.y]) { vis1[t.x][t.y] = 1; Q.push(t); } } } } int mx = -INF; for(int i = 0; i < f.size(); i++) { mx = max(mx, d[f[i].x][f[i].y]); } return mx; } set<int> s1, s2; vector<node> G; int bfs(int x1, int y1, int x2, int y2) { queue<node> Q; for(int i = 0; i <= n; i++) for(int j = 0; j <= m; j++) d[i][j] = INF; vector<node> f; d[x1][y1] = d[x2][y2] = 0; Q.push(node(x1, y1)); Q.push(node(x2, y2)); mem(vis1, 0); vis1[x1][y1] = 1, vis1[x2][y2] = 1; while(!Q.empty()) { node u = Q.front(); Q.pop(); f.push_back(u); vis1[u.x][u.y] = 0; for(int i = 0; i < 4; i++) { node t = node(0, 0); t.x = u.x + dis[i][0]; t.y = u.y + dis[i][1]; if(t.x < 0 || t.x >= n || t.y < 0 || t.y >= m || str[t.x][t.y] == '.') continue; if(d[t.x][t.y] > d[u.x][u.y] + 1) { d[t.x][t.y] = d[u.x][u.y] + 1; if(!vis1[t.x][t.y]) { vis1[t.x][t.y] = 1; Q.push(t); } } } } int mx = -INF; for(int i = 0; i < f.size(); i++) { mx = max(mx, d[f[i].x][f[i].y]); } return mx; } int main() { int T, kase = 0; rd(T); while(T--) { s1.clear(); s2.clear(); mem(vis, 0); int cnt = 0; G.clear(); rd(n), rd(m); for(int i = 0; i < n; i++) { cin >> str[i]; for(int j = 0; j < m; j++) if(str[i][j] == '#') G.push_back(node(i, j)); } // cout << 11111 <<endl; for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) if(!vis[i][j] && str[i][j] == '#') { cnt++; vis[i][j] = cnt; dfs(i, j, cnt); } //cout << cnt << endl; printf("Case %d: ", ++kase); if(cnt > 2) { printf("-1 "); continue; } if(cnt == 0) { printf("0 "); continue; } if(cnt == 1) { for(int i = 0; i < G.size(); i++) { node u1 = G[i]; for(int j = i + 1; j < G.size(); j++) { node u2 = G[j]; s1.insert(bfs(u1.x, u1.y, u2.x, u2.y)); } } set<int>::iterator it; it = s1.begin(); cout << *it << endl; continue; } for(int i = 0; i < n; i++) { for(int j = 0; j < m; j++) { if(str[i][j] == '#') { if(vis[i][j] == 1) s1.insert(spfa(i, j)); else s2.insert(spfa(i, j)); } } } set<int>::iterator it, is; it = s1.begin(); is = s2.begin(); int mxx = max(*it, *is); cout << mxx << endl; } return 0; }