hdu 3333 Turing Tree

题目链接

给n个数， m个询问， 每次询问输出区间内的数的和， 相同的数只计算一次。

数组里的数是>-1e9 <1e9， 可以把它离散以后用莫队搞...

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 3e5+5;
int a[maxn], b[maxn], c[maxn];
ll ans[100005];
struct node
{
    int l, r, id, block;
    bool operator < (node a)const
    {
        if(block == a.block)
            return r<a.r;
        return block<a.block;
    }
}q[100005];
int main()
{
    int t, n, m;
    cin>>t;
    while(t--) {
        scanf("%d", &n);
        mem(c);
        for(int i = 1; i<=n; i++) {
            scanf("%d", &a[i]);
            b[i-1] = a[i];
        }
        sort(b, b+n);
        int cnt = unique(b, b+n)-b;
        for(int i = 1; i<=n; i++) {
            a[i] = lower_bound(b, b+cnt, a[i])-b+1;
        }
        int BLOCK = max(1.0, sqrt(cnt*1.0));
        cin>>m;
        for(int i = 0; i<m; i++) {
            scanf("%d%d", &q[i].l, &q[i].r);
            q[i].block = q[i].l/BLOCK;
            q[i].id = i;
        }
        sort(q, q+m);
        ll sum = 0;
        for(int i = q[0].l; i<=q[0].r; i++) {
            if(!c[a[i]]) {
                sum += b[a[i]-1];
            }
            c[a[i]]++;
        }
        ans[q[0].id] = sum;
        for(int i = 1; i<m; i++) {
            if(q[i].l<q[i-1].l) {
                for(int j = q[i-1].l-1; j>=q[i].l; j--) {
                    if(!c[a[j]]) {
                        sum += b[a[j]-1];
                    }
                    c[a[j]]++;
                }
            } else {
                for(int j = q[i-1].l; j<q[i].l; j++) {
                    if(c[a[j]]==1) {
                        sum -= b[a[j]-1];
                    }
                    c[a[j]]--;
                }
            }
            if(q[i].r>q[i-1].r) {
                for(int j = q[i-1].r+1; j<=q[i].r; j++) {
                    if(!c[a[j]]) {
                        sum += b[a[j]-1];
                    }
                    c[a[j]]++;
                }
            } else {
                for(int j = q[i-1].r; j>q[i].r; j--) {
                    if(c[a[j]] == 1) {
                        sum -= b[a[j]-1];
                    }
                    c[a[j]]--;
                }
            }
            ans[q[i].id] = sum;
        }
        for(int i = 0; i<m; i++)
            printf("%I64d
", ans[i]);
    }
    return 0;
}