題意：有兩種操作：(1)插入線段，第i次插入的線段左邊界為Li，長度為i (2)刪除線段，刪除第x次插入的線段。每次插入線段之前詢問有多少條線段被它覆蓋。

思路：由于插入的線段長度是遞增的，所以第i次插入的線段的長度比以前插入的所有線段都要長，從以前插入的線段里面任取一條，考慮其與當(dāng)前線段的位置關(guān)系，主要有以下三種：

圖中，下方表示當(dāng)前線段。注意到，只有被當(dāng)前線段覆蓋的線段，它的左右邊界和當(dāng)前線段的左右邊界的相對位置是不同的。也就是說，查詢有多少個線段的右端點小于等于該線段右端點，再查詢有多少條線段左端點小于該線段的左端點， 兩者之差就是答案。因為不符合要求的線段同時進入了前者和后者，而符合要求的答案只進入了前者。

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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_JUDGEfreopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGEint 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:\n", ++ 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\n", rta.query(rid) - lta.query(lid - 1));lta.update(lid, 1);rta.update(rid, 1);}}}return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/jklongint/p/4723997.html

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