hdu 3874（树状数组+离线算法）

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解题思路：这道题和之前的题一样，查找[l,r]区间内不重复数字的和。可以利用离线算法，实际上离线算法为了解决在查找时出现的矛盾，因为每次询问的区间大小不同，如果单独处理的话可能会对之后的查询有影响，所以离线算法帮助我们把要查询的区间先按照右端点进行排序，因为在处理更靠右的区间时，比它小的区间肯定是已经得到的不重复的数了，所以不会影响到后面的查找。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int M = 200005;const int N = 50005;struct Query{int l,r,id;bool operator < (Query a){return r < a.r;}}q[M];int n,m,size;int tree[N<<1],arr[N],tmp[N],last[N],ans[M];int lowbit(int x){return x & -x;}void update(int x,int d){for(int i = x; i <= n; i += lowbit(i))tree[i] += d;}int getsum(int x){int sum = 0;for(int i = x; i > 0; i -= lowbit(i))sum += tree[i];return sum;}int bisearch(int val){int l = 1, r = size, mid;while(l <= r){mid = (l + r) >> 1;if(tmp[mid] == val) return mid;else if(tmp[mid] > val)r = mid - 1;else l = mid + 1;}}int main(){int t;cin >> t;while(t--){cin >> n;for(int i = 1; i <= n; i++){cin >> arr[i];tmp[i] = arr[i];}cin >> m;for(int i = 1; i <= m; i++){cin >> q[i].l >> q[i].r;q[i].id = i;}sort(tmp+1,tmp+1+n);sort(q+1,q+1+m);//离散化size = 1;for(int i = 2; i <= n; i++)if(tmp[i] != tmp[i-1])tmp[++size] = tmp[i];memset(tree,0,sizeof(tree));memset(last,0,sizeof(last));int idx = 1;for(int i = 1; i <= n; i++){int x = bisearch(arr[i]);if(last[x])update(last[x],-arr[i]);update(i,arr[i]);last[x] = i;while(q[idx].r == i && idx <= m){ans[q[idx].id] = getsum(q[idx].r) - getsum(q[idx].l - 1);idx++;}}for(int i = 1; i <= m; i++)cout << ans[i] << endl;}return 0;}

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