标签:style blog os io for ar 问题 div
题意: 给你一个数列,然后有n个查询,问你给定区间中不同数字的和是多少。
思路还是比较难想的,起码对于蒟蒻我来说。
将区间按照先右端点,后左端点从小到大排序之后,对于每个查询,我只要维护每个数字出现的最后一次就可以了(这个结论稍微想一下就可以证明是正确的)。
然后就是简单的点更新,区间求和问题了~
#include <cstdio> #include <cstring> #include <iostream> #include <map> #include <set> #include <vector> #include <string> #include <queue> #include <deque> #include <bitset> #include <list> #include <cstdlib> #include <climits> #include <cmath> #include <ctime> #include <algorithm> #include <stack> #include <sstream> #include <numeric> #include <fstream> #include <functional> using namespace std; #define MP make_pair #define PB push_back typedef long long LL; typedef unsigned long long ULL; typedef vector<int> VI; typedef pair<int,int> pii; const int INF = INT_MAX / 3; const double eps = 1e-8; const LL LINF = 1e17; const double DINF = 1e60; const int maxn = 3e4 + 10; struct Seg { int l,r,id; LL ans; Seg(int l,int r,int id): l(l),r(r),id(id) {} }; int n,val[maxn],last[maxn],m; VI num; vector<Seg> query; LL C[maxn]; bool ext[maxn]; bool cmp(const Seg &a,const Seg &b) { if(a.r == b.r) return a.l < b.l; return a.r < b.r; } bool cmp1(const Seg &a,const Seg &b) { return a.id < b.id; } inline int lowbit(int x) { return x & (-x); } void addv(int pos,LL v) { while(pos <= n) { C[pos] += v; pos += lowbit(pos); } } LL ask_(int pos) { LL ret = 0; while(pos > 0) { ret += C[pos]; pos -= lowbit(pos); } return ret; } LL ask(int l,int r) { return ask_(r) - ask_(l - 1); } int getID(int Val) { return lower_bound(num.begin(),num.end(),Val) - num.begin() + 1; } void solve() { memset(C,0,sizeof(C)); memset(last,-1,sizeof(last)); int npos = 1; for(int i = 0;i < m;i++) { Seg &now = query[i]; while(npos <= now.r) { int nowval = getID(val[npos]); if(last[nowval] != -1) addv(last[nowval],-val[npos]); last[nowval] = npos; addv(npos,val[npos]); npos++; } now.ans = ask(now.l,now.r); } } int main() { int T; scanf("%d",&T); while(T--) { query.clear(); num.clear(); int tmp; scanf("%d",&n); for(int i = 1;i <= n;i++) { scanf("%d",&val[i]); num.PB(val[i]); } sort(num.begin(),num.end()); num.erase(unique(num.begin(),num.end()),num.end()); scanf("%d",&m); for(int i = 0;i < m;i++) { int l,r; scanf("%d%d",&l,&r); query.PB(Seg(l,r,i)); } sort(query.begin(),query.end(),cmp); solve(); sort(query.begin(),query.end(),cmp1); for(int i = 0;i < m;i++) cout << query[i].ans << endl; } return 0; }
HDU 3333 Turing Tree 树状数组 离线查询,布布扣,bubuko.com
HDU 3333 Turing Tree 树状数组 离线查询
标签:style blog os io for ar 问题 div
原文地址:http://www.cnblogs.com/rolight/p/3925437.html