HDU 3333 Turing Tree

Problem Description

After inventing Turing Tree, 3xian always felt boring when solving problems about intervals, because Turing Tree could easily have the solution. As well, wily 3xian made lots of new problems about intervals. So, today, this sick thing happens again...

Now given a sequence of N numbers A1, A2, ..., AN and a number of Queries(i, j) (1≤i≤j≤N). For each Query(i, j), you are to caculate the sum of distinct values in the subsequence Ai, Ai+1, ..., Aj.

Input

The first line is an integer T (1 ≤ T ≤ 10), indecating the number of testcases below.
For each case, the input format will be like this:
* Line 1: N (1 ≤ N ≤ 30,000).
* Line 2: N integers A1, A2, ..., AN (0 ≤ Ai ≤ 1,000,000,000).
* Line 3: Q (1 ≤ Q ≤ 100,000), the number of Queries.
* Next Q lines: each line contains 2 integers i, j representing a Query (1 ≤ i ≤ j ≤ N).

Output

For each Query, print the sum of distinct values of the specified subsequence in one line.

Sample Input

2
3
1 1 4
2
1 2
2 3
5
1 1 2 1 3
3
1 5
2 4
3 5

Sample Output

1
5
6
3
6

此题和 hdu 4358 Boring counting 类似，可以说是hdu4358的简化版，
具体可以看http://blog.csdn.net/jtjy568805874/article/details/43925895

树状数组的解法有两种：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>  
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<map>
using namespace std;
const int maxn = 100005;
const int low(int x){ return x&-x; }
map<int, int> M, N;
int T, n, k, a[maxn], p[maxn], m;
long long f[maxn], ans[maxn];

struct abc
{
	int a, b, c;
	bool operator <(const abc&x)
	{
		if (x.c == c) return b < x.b;
		return c < x.c;
	}
}r[maxn];


void add(int x, long long y)
{
	for (int i = x; i <= n; i += low(i)) f[i] += y;
}

long long sum(int x)
{
	long long tot = 0;
	for (; x; x -= low(x)) tot += f[x];
	return tot;
}

int main()
{
	scanf("%d", &T);
	while (T--)
	{
		cin >> n;
		int x, y = 0;
		M.clear();	N.clear();
		memset(f, 0, sizeof(f));
		memset(p, 0, sizeof(p));
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &x);
			if (M.count(x)) a[i] = M[x]; else
			{
				a[i] = M[x] = ++y;
				N[y] = x;
			}
		}

		scanf("%d", &m);
		for (int i = 1; i <= m; r[i].a = i++) scanf("%d%d", &r[i].b, &r[i].c);
		sort(r + 1, r + m + 1);

		for (int i = 1, j = 1; i <= n&&j <= m; i++)
		{
			if (p[a[i]]) add(p[a[i]], -N[a[i]]);
			p[a[i]] = i;
			add(p[a[i]], N[a[i]]);
			for (; j <= m&&r[j].c == i; j++) ans[r[j].a] = sum(r[j].c) - sum(r[j].b - 1);
		}
	
		for (int i = 1; i <= m; i++) cout << ans[i] << endl; 
	}
	return 0;
}
第二种，同hdu4358.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>  
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<map>
using namespace std;
const int maxn = 100005;
const int low(int x){ return x&-x; }
map<int, int> M, N;
int T, n, k, a[maxn], p[maxn], m;
long long f[maxn], ans[maxn];

struct abc
{
	int a, b, c;
	bool operator <(const abc&x)
	{
		if (x.c == c) return b < x.b;
		return c < x.c;
	}
}r[maxn];


void add(int x, long long y)
{
	for (int i = x; i <= n; i += low(i)) f[i] += y;
}

long long sum(int x)
{
	long long tot = 0;
	for (; x; x -= low(x)) tot += f[x];
	return tot;
}

int main()
{
	scanf("%d", &T);
	while (T--)
	{
		cin >> n;
		int x, y = 0;
		M.clear();	N.clear();
		memset(f, 0, sizeof(f));
		memset(p, 0, sizeof(p));
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &x);
			if (M.count(x)) a[i] = M[x]; else
			{
				a[i] = M[x] = ++y;
				N[y] = x;
			}
		}

		scanf("%d", &m);
		for (int i = 1; i <= m; r[i].a = i++) scanf("%d%d", &r[i].b, &r[i].c);
		sort(r + 1, r + m + 1);

		for (int i = 1, j = 1; i <= n&&j <= m; i++)
		{
			add(p[a[i]] + 1, N[a[i]]);
			p[a[i]] = i;
			add(p[a[i]] + 1, -N[a[i]]);
			for (; j <= m&&r[j].c == i; j++) ans[r[j].a] = sum(r[j].b);
		}

		for (int i = 1; i <= m; i++) cout << ans[i] << endl;
	}
	return 0;
}
zkw线段树

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>  
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<map>
using namespace std;
const int maxn = 100005;
const int low(int x){ return x&-x; }
map<int, int> M, N;
int T, n, k, a[maxn], p[maxn], m, u;
long long f[maxn], ans[maxn];

struct abc
{
	int a, b, c;
	bool operator <(const abc&x)
	{
		if (x.c == c) return b < x.b;
		return c < x.c;
	}
}r[maxn];


void change(int x, int y)
{
	x += u;		f[x] = y;
	for (x >>= 1; x; x >>= 1) f[x] = f[x + x] + f[x + x + 1];
}

long long find(int l, int r)
{
	long long tot = 0;
	for (l += u - 1, r += u + 1; l ^ r ^ 1; l >>= 1, r >>= 1)
	{
		if (~l & 1) tot += f[l ^ 1];
		if ( r & 1) tot += f[r ^ 1];
	}
	return tot;
}

int main()
{
	scanf("%d", &T);
	while (T--)
	{
		cin >> n;
		for (u = 1; u < n; u += u);
		int x, y = 0;
		M.clear();	N.clear();
		memset(f, 0, sizeof(f));
		memset(p, 0, sizeof(p));
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &x);
			if (M.count(x)) a[i] = M[x]; else
			{
				a[i] = M[x] = ++y;
				N[y] = x;
			}
		}

		scanf("%d", &m);
		for (int i = 1; i <= m; r[i].a = i++) scanf("%d%d", &r[i].b, &r[i].c);
		sort(r + 1, r + m + 1);

		for (int i = 1, j = 1; i <= n&&j <= m; i++)
		{
			if (p[a[i]]) change(p[a[i]], 0);
			p[a[i]] = i;
			change(p[a[i]], N[a[i]]);
			for (; j <= m&&r[j].c == i; j++) ans[r[j].a] = find(r[j].b, r[j].c);
		}

		for (int i = 1; i <= m; i++) cout << ans[i] << endl;
	}
	return 0;
}