[SCOI2009]生日礼物

[SCOI2009]生日礼物

// Problem: 丢手绢// Contest: NowCoder// URL: Memory Limit: 524288 MB// Time Limit: 2000 ms// 2022-03-09 14:00:09// // Powered by CP Editor (namespace std;#define rep(i,l,r) for(int i=(l);i<=(r);i++)#define per(i,l,r) for(int i=(l);i>=(r);i--)#define ll long long#define mset(s,t) memset(s,t,sizeof(t))#define mcpy(s,t) memcpy(s,t,sizeof(t))#define fi first#define se second#define pb push_back#define all(x) (x).begin(),(x).end()#define SZ(x) ((int)(x).size())#define mp make_pairtypedef pair pii;typedef pair pll;typedef vector vi; typedef vector Vll; typedef vector > vpii;typedef vector > vpll; const ll mod = 1e9 + 7;//const ll mod = 998244353;const double pi = acos(-1.0);inline ll qmi (ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans = ans * a%mod; a = a * a %mod; b >>= 1; } return ans;}inline int read () { int x = 0, f = 0; char ch = getchar(); while (!isdigit(ch)) f |= (ch=='-'),ch= getchar(); while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar(); return f?-x:x;}template void print(T x) { if (x < 0) putchar('-'), x = -x; if (x >= 10) print(x/10); putchar(x % 10 + '0');}inline ll sub (ll a, ll b) { return ((a - b ) %mod + mod) %mod;}inline ll add (ll a, ll b) { return (a + b) %mod;}inline ll inv (ll a) { return qmi(a, mod - 2);}int num[110];int n, m;vector a;int kind;void init () { cin >> n >> m; for (int i =1; i<= m; i ++) { int c;cin >> c; for (int j = 1; j <= c; j ++) { int x; cin >> x; a.pb({x, i}); } } sort(all(a));}void stat (int p, bool f) { if (f) if ((++num[a[p].se]) == 1) kind ++; if (!f) if ((--num[a[p].se]) == 0) kind --;}void solve() { init(); int l = 0,r = 0; stat(0, 1);// for (auto t : a) {// cout << t.fi << " " << t.se << endl;// } int ans = 0x3f3f3f3f; while (l <= r) { while (kind < m && r+ 1 > t; while (t --) solve(); return 0;}

题意:给你一个坐标。每个位置上面有彩珠.问要求选到k种颜色的珠子。满足要求的最小距离是多少？

通过单调队列维护。逐渐向外扩。对于kind = m的判断距离取最小值通过把加减封装成函数

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