HDU - 5997 树状数组+set
和之前一道省选题目很像;
(梦幻布丁);
我们维护的时候用树状数组维护,在断电处打上标记;
合并的时候小的合并到大的里面;
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<functional> #include<sstream> //#include<cctype> //#pragma GCC optimize(2) using namespace std; #define maxn 200005 #define inf 0x7fffffff //#define INF 1e18 #define rdint(x) scanf("%d",&x) #define rdllt(x) scanf("%lld",&x) #define rdult(x) scanf("%lu",&x) #define rdlf(x) scanf("%lf",&x) #define rdstr(x) scanf("%s",x) typedef long long ll; typedef unsigned long long ull; typedef unsigned int U; #define ms(x) memset((x),0,sizeof(x)) const long long int mod = 1e9; #define Mod 1000000000 #define sq(x) (x)*(x) #define eps 1e-5 typedef pair<int, int> pii; #define pi acos(-1.0) //const int N = 1005; #define REP(i,n) for(int i=0;i<(n);i++) typedef pair<int, int> pii; inline int rd() { int x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x; } ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); } int sqr(int x) { return x * x; } /*ll ans; ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans; } */ int T; int n, q; set<int>st[1000005]; int a[maxn]; int c[maxn]; int fa[1000005]; void add(int x, int val) { while (x <= n) { c[x] += val; x += x & -x; } } int query(int x) { int ans = 0; while (x > 0) { ans += c[x]; x -= x & -x; } return ans; } void upd(int x, int y) { for (auto cnt : st[x]) { if (a[cnt - 1] == y)add(cnt, -1); if (a[cnt + 1] == y)add(cnt + 1, -1); st[y].insert(cnt); } for (auto cnt : st[x])a[cnt] = y; st[x].clear(); } int main() { //ios::sync_with_stdio(0); T = rd(); while (T--) { rdint(n); rdint(q); for (int i = 0; i <=1000002; i++)st[i].clear(); ms(fa); ms(c); ms(a); for (int i = 1; i <= n; i++) { rdint(a[i]); } // st[a[1]].insert(1); for (int i = 1; i <= n; i++) { fa[a[i]] = a[i]; if (a[i] != a[i - 1])add(i, 1); st[a[i]].insert(i); } while (q--) { int opt; rdint(opt); if (opt == 1) { int x, y; rdint(x); rdint(y); if (x == y)continue; if (st[fa[x]].size() > st[fa[y]].size())swap(fa[x], fa[y]); x = fa[x]; y = fa[y]; upd(x, y); } else { int x, y; rdint(x); rdint(y); int ans = query(y) - query(x - 1) + (a[x] == a[x - 1]); printf("%d\n", ans); } } } return 0; }
EPFL - Fighting