1204C - Anna, Svyatoslav and Maps
1204C - Anna, Svyatoslav and Maps
看别人题解才知道题目是怎么回事。。。
#include <iostream> #include <vector> #include <algorithm> #include <string> #include <set> #include <queue> #include <map> #include <sstream> #include <cstdio> #include <cstring> #include <numeric> #include <cmath> #include <iomanip> #include <deque> #include <bitset> //#include <unordered_set> //#include <unordered_map> #define ll long long #define pii pair<int, int> #define rep(i,a,b) for(ll i=a;i<=b;i++) #define dec(i,a,b) for(ll i=a;i>=b;i--) #define forn(i, n) for(ll i = 0; i < int(n); i++) using namespace std; int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } }; const long long INF = 0x7f7f7f7f7f7f7f7f; const int inf = 0x3f3f3f3f; const double pi = 3.14159265358979323846; const double eps = 1e-6; const int mod = 998244353; const int N = 1e6 + 5; //if(x<0 || x>=r || y<0 || y>=c) inline ll read() { ll x = 0; bool f = true; char c = getchar(); while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); } while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar(); return f ? x : -x; } ll gcd(ll m, ll n) { return n == 0 ? m : gcd(n, m % n); } ll lcm(ll m, ll n) { return m * n / gcd(m, n); } bool prime(int x) { if (x < 2) return false; for (int i = 2; i * i <= x; ++i) { if (x % i == 0) return false; } return true; } inline int qpow(int x, ll n) { int r = 1; while (n > 0) { if (n & 1) r = 1ll * r * x % mod; n >>= 1; x = 1ll * x * x % mod; } return r; } inline int add(int x, int y) { return ((x%mod)+(y%mod))%mod; } inline int sub(int x, int y) { x -= y; return x < 0 ? x += mod : x; } inline int mul(int x, int y) { return (1ll * (x %mod) * (y % mod))%mod; } inline int Inv(int x) { return qpow(x, mod - 2); } int main() { int n; cin >> n; vector<vector<int>> g(n+1,vector<int>(n+1,inf)); rep(i, 1, n) { string s; cin >> s; forn(j, n) { if (s[j] == '1') { g[i][j + 1] = 1; } } } rep(i, 0, n) g[i][i] = 0; int len; cin >> len; vector<int> a(len + 1); rep(i, 1, len) cin >> a[i]; rep(i, 1, n) { rep(j, 1, n) { rep(k, 1, n) { g[j][k] = min(g[j][i] + g[i][k], g[j][k]); } } } vector<int> res; res.push_back(a[1]); int dis = 0; rep(i, 2, len) { dis += g[a[i - 1]][a[i]]; if (dis > g[res.back()][a[i]]) { res.push_back(a[i - 1]); dis = g[res.back()][a[i]]; } } res.push_back(a[len]); cout << res.size() << endl; rep(i, 0, res.size()-1) cout << res[i] << " "; return 0; }