[hdu5372 Segment Game]树状数组

题意：有两种操作：(1)插入线段，第i次插入的线段左边界为Li，长度为i (2)删除线段，删除第x次插入的线段。每次插入线段之前询问有多少条线段被它覆盖。

思路：由于插入的线段长度是递增的，所以第i次插入的线段的长度比以前插入的所有线段都要长，从以前插入的线段里面任取一条，考虑其与当前线段的位置关系，主要有以下三种：

图中，下方表示当前线段。注意到，只有被当前线段覆盖的线段，它的左右边界和当前线段的左右边界的相对位置是不同的。也就是说，查询有多少个线段的右端点小于等于该线段右端点，再查询有多少条线段左端点小于该线段的左端点， 两者之差就是答案。因为不符合要求的线段同时进入了前者和后者，而符合要求的答案只进入了前者。

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//#pragma comment(linker, "/STACK:1024000000") #include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; //#ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} //#endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} template<typename T> void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} template<typename T> void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} #if 0 const double PI = acos(-1.0); const int INF = 1e9 + 7; const double EPS = 1e-8; #endif /* -------------------------------------------------------------------------------- */ const int maxn = 4e5 + 7; int u[maxn], v[maxn], w[maxn], tot; pii rem[maxn]; class TA { int C[maxn], n; inline int lowbit(int x) { return x & -x; } public: void init(int n) { this->n = n; for (int i = 1; i <= n; i ++) C[i] = 0; } void update(int p, int x) { while (p <= n) { C[p] += x; p += lowbit(p); } } int query(int p) { int ans = 0; while (p) { ans += C[p]; p -= lowbit(p); } return ans; } }; TA lta, rta; inline int find(int x) { return lower_bound(w, w + tot, x) - w + 1; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int cas = 0, n; while (cin >> n) { tot = 0; int len = 1; for (int i = 0; i < n; i ++) { scanf("%d%d", u + i, v + i); if (u[i] == 0) { rem[len] = mp(v[i], v[i] + len); w[tot ++] = v[i]; w[tot ++] = v[i] + len ++; } } sort(w, w + tot); tot = unique(w, w + tot) - w; lta.init(tot); rta.init(tot); printf("Case #%d: ", ++ cas); len = 1; for (int i = 0; i < n; i ++) { if(u[i] == 1) { pii buf = rem[v[i]]; int lid = find(buf.X), rid = find(buf.Y); lta.update(lid, - 1); rta.update(rid, - 1); } else { int lid = find(v[i]), rid = find(v[i] + len ++); printf("%d ", rta.query(rid) - lta.query(lid - 1)); lta.update(lid, 1); rta.update(rid, 1); } } } return 0; }