HDU3874 Necklace 树状数组+离线处理

这题是要求一段区间内的不重复的数字之和。我们通过对询问区间的右端点进行排序，然后记录每一数字的上一次的出现的位置，由于询问都是不回溯的那么就可以线性的更新了。

代码如下：

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 50005
using namespace std;

typedef long long int Int64;

int N, M, seq[MAXN], last[1000005];

Int64 ans[200005], c[MAXN];

int lowbit(int x)
{
    return x & -x;
}

void modify(int x, int val)
{
    for (int i = x; i <= N; i += lowbit(i)) {
        c[i] += val;    
    }
}

Int64 sum(int x)
{
    Int64 ret = 0;
    for (int i = x; i > 0; i -= lowbit(i)) {
        ret += c[i];
    }
    return ret;
}

struct Node
{
    int a, b, no;
    bool operator < (Node temp) const
    {
        return b < temp.b;    
    }
}q[200005];

int main()
{
    int T, ptr;
    scanf("%d", &T);
    while (T--) {
        memset(last, 0, sizeof (last));
        memset(c, 0, sizeof (c));
        scanf("%d", &N);
        ptr = 1;
        for (int i = 1; i <= N; ++i) {
            scanf("%d", &seq[i]);
        }
        scanf("%d", &M);
        for (int i = 1; i <= M; ++i) {
            scanf("%d %d", &q[i].a, &q[i].b);
            if (q[i].a > q[i].b) {
                int t = q[i].a;
                 q[i].a = q[i].b;
                q[i].b = t;
            }
            q[i].no = i;
        }
        sort(q+1, q+M+1);
        for (int i = 1; i <= M; ++i) {
            while (ptr <= q[i].b) {
                if (last[seq[ptr]]) {
                    modify(last[seq[ptr]], -seq[ptr]);
                }
                last[seq[ptr]] = ptr;
                modify(ptr, seq[ptr]);
                ++ptr;
            }
            ans[q[i].no] = sum(q[i].b) - sum(q[i].a-1);
        }
        for (int i = 1; i <= M; ++i) {
            printf("%I64d\n", ans[i]);
        }
    }
    return 0;    
}