hdu 3333 Turing Tree(线段树)

题目链接：hdu 3333 Turing Tree

题目大意：给定一个长度为N的序列。有M次查询，每次查询l。r之间元素的总和，同样元素仅仅算一次。

解题思路：涨姿势了，线段树的一种题型。离线操作，将查询依照右区间排序，每次考虑一个询问。将mv ~ r的点所有标记为存在。而且对于每一个位置i。假设A[i]在前面已经出现过了。那么将前面的那个位置减掉A[i]。当前位置加入A[i]，这样做维护了每一个数尽量做，那么碰到查询。用sum[r] - sum[l-1]就可以。

#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>

using namespace std;

typedef long long ll;

const int maxn = 30000;

int N, Q;
ll A[maxn+5], ans[100005];
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;
}