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，相同元素只算一次。

解题思路：涨姿势了，线段树的一种题型，离线操作，将查询按照右区间排序，每次考虑一个询问，将mv ~ r的点全部标记为存在，并且对于每个位置i，如果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]，当前位置添加A[i]，这样做维护了每个数尽量做，那么碰到查询，用sum[r] - sum[l-1]即可。

#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;
}