【CF1000C】Covered Points Count（离散化+差分）

【【CF1000C】Covered Points Count（离散化+差分）】的更多相关文章

cf1000C Covered Points Count (差分+map)

考虑如果数字范围没有这么大的话,直接做一个差分数组就可以了但现在变大了所以要用一个map来维护 #include<bits/stdc++.h> #define pa pair<int,int> #define CLR(a,x) memset(a,x,sizeof(a)) using namespace std; typedef long long ll; ; inline ll rd(){ ll x=;; ;c=getchar();} +c-',c=getchar(); ret…

codeforces 1000C - Covered Points Count 【差分】

题目:戳这里题意:给出n个线段,问被1~n个线段覆盖的点分别有多少. 解题思路: 这题很容易想到排序后维护每个端点被覆盖的线段数,关键是端点值不好处理.比较好的做法是用差分的思想,把闭区间的线段改为前闭后开,同时在求总点数的时候,也按前闭后开的区间来求,这样就巧妙避开了两个端点之间的讨论,只用维护好一个端点就行了. 代码比文字更容易理解: 1 #include <bits/stdc++.h> 2 #define lowbit(x) x&-x; 3 typedef long long…

C - Covered Points Count CodeForces - 1000C （差分，离散化，统计）

C - Covered Points Count CodeForces - 1000C You are given nn segments on a coordinate line; each endpoint of every segment has integer coordinates. Some segments can degenerate to points. Segments can intersect with each other, be nested in each othe…

Covered Points Count CF1000C 思维前缀和贪心

Covered Points Count time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output You are given nn segments on a coordinate line; each endpoint of every segment has integer coordinates. Some segments…

Educational Codeforces Round 46 C - Covered Points Count

C - Covered Points Count emmm 好像是先离散化一下注意 R需要+1 这样可以确定端点 emmm 扫描线?瞎搞一下? #include<bits/stdc++.h> using namespace std; #define maxn 4000005 #define LL long long LL a[maxn],b[maxn],ll[maxn],rr[maxn],c[maxn]; LL x[maxn],y[maxn]; vector<LL >q; int…

Covered Points Count（思维题）

C. Covered Points Count time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output You are given nn segments on a coordinate line; each endpoint of every segment has integer coordinates. Some segment…