hdu 3874 Necklace(线段树)
2014-10-11 17:26
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题目链接:hdu 3874 Necklace
题目大意:和hdu 3333 Turing Tree是一样的。不解释。
题目大意:和hdu 3333 Turing Tree是一样的。不解释。
#include <cstdio> #include <cstring> #include <vector> #include <map> #include <algorithm> using namespace std; typedef long long ll; const int maxn = 50000; int N, Q; ll A[maxn+5], ans[200005]; map<ll, int> G; #define lson(x) ((x)<<1) #define rson(x) (((x)<<1)|1) int lc[maxn << 2], rc[maxn << 2]; ll s[maxn << 2]; inline void pushup (int u) { s[u] = s[lson(u)] + s[rson(u)]; } void build (int u, int l, int r) { lc[u] = l; rc[u] = r; s[u] = 0; if (l == r) return; int mid = (lc[u] + rc[u]) / 2; build(lson(u), l, mid); build(rson(u), mid + 1, r); pushup(u); } void modify(int u, int x, ll d) { if (x == lc[u] && rc[u] == x) { s[u] += d; return; } int mid = (lc[u] + rc[u]) / 2; if (x <= mid) modify(lson(u), x, d); else modify(rson(u), x, d); pushup(u); } ll query(int u, int l, int r) { if (l <= lc[u] && rc[u] <= r) return s[u]; ll ret = 0; int mid = (lc[u] + rc[u]) / 2; if (l <= mid) ret += query(lson(u), l, r); if (r > mid) ret += query(rson(u), l, r); pushup(u); return ret; } struct Seg { int l, r, id; Seg (int l = 0, int r = 0, int id = 0) { this->l = l; this->r = r; this->id = id; } friend bool operator < (const Seg& a, const Seg& b) { return a.r < b.r; } }; vector<Seg> vec; void init () { int l, r; G.clear(); vec.clear(); scanf("%d", &N); for (int i = 1; i <= N; i++) scanf("%I64d", &A[i]); scanf("%d", &Q); for (int i = 1; i <= Q; i++) { scanf("%d%d", &l, &r); vec.push_back(Seg(l, r, i)); } sort(vec.begin(), vec.end()); } void solve () { build (1, 0, N); int k = 0; for (int i = 0; i < Q; i++) { for ( ; k <= vec[i].r; k++) { if (G[A[k]]) modify(1, G[A[k]], -A[k]); G[A[k]] = k; modify(1, k, A[k]); } ans[vec[i].id] = query(1, vec[i].l, vec[i].r); } for (int i = 1; i <= Q; i++) printf("%I64d\n", ans[i]); } int main () { int cas; scanf("%d", &cas); while (cas--) { init(); solve(); } return 0; }
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