题目链接:https://vjudge.net/problem/POJ-2443 Set Operation Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 3554   Accepted: 1477 Description You are given N sets, the i-th set (represent by S(i)) have C(i) element (Here "set" isn't enti…

模板题.S[i][j]表示i是否存在于第j个集合里.妈蛋poj差点打成poi(波兰无关)是不是没救了. #include<cstdio> #include<bitset> using namespace std; int n,m,x,y; bitset<1000>S[10001]; int main() { scanf("%d",&n); for(int i=0;i<n;++i) { scanf("%d",&…

You are given N sets, the i-th set (represent by S(i)) have C(i) element (Here "set" isn't entirely the same as the "set" defined in mathematics, and a set may contain two same element). Every element in a set is represented by a posit…

传送门 题意:给出n个集合(n<=1000),每个集合中最多有10000个数,每个数的范围为1~10000,给出q次询问(q<=200000),每次给出两个数u,v判断是否有一个集合中同时含有u,v两个数 枚举每一个集合,看看是否同时又u和v,显然超时 用bitset维护每一个数所在集合,求解的时候直接u & v即可 #include <cstdio> #include <bitset> using namespace std; int n, m; bitset…

题目链接:http://poj.org/problem?id=2443 题目大意:给你N个集合,每个集合里有若干个数.M个查询,每个查询有a,b两个数.问是否存在一个集合同时包含a,b这两个数.若存在则输出Yes,否则为No. 康神竟然一下子就想出来了.. 思路:统计每个数在哪个集合出现过,用bitset记录下来.然后对于a,b,则把两个bitset取与. 如果得到空集就是No,否则就是Yes. #include <cstdio> #include <bitset> #includ…

Description You are given N sets, the i-th set (represent by S(i)) have C(i) element (Here "set" isn't entirely the same as the "set" defined in mathematics, and a set may contain two same element). Every element in a set is represente…

POJ2443 Set Operation Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 2679   Accepted: 1050 Description You are given N sets, the i-th set (represent by S(i)) have C(i) element (Here "set" isn't entirely the same as the "set&qu…

学了下bitset用法,从网上找的一些bitset用法,并从中调出一些常用的用法. 构造函数bitset<n> b; b有n位,每位都为0.参数n可以为一个表达式.如bitset<5> b0;则"b0"为"00000"; bitset<n> b(unsigned long u); b有n位,并用u赋值;如果u超过n位,则顶端被截除如:bitset<5>b0(5);则"b0"为"00101&…

Bryce1010模板 /* 题意:给出n个集合(n<=1000),每个集合中最多有10000个数, 每个数的范围为1~10000,给出q次询问(q<=200000), 每次给出两个数u,v判断是否有一个集合中同时含有u,v两个数 */ #include<iostream> #include<bitset> #include<stdio.h> using namespace std; bitset<1001>b[10010]; int main(…

---------------------------------------------------------------------------- 一题题目: 一题题解: 这个题目哪来入门再好不过了,支老板之前没有接触过这个东西,然后一点即通:就是把一个int(32位)拆成32个只放0或1的位置,然后这32个的单点操作或者32个一起操作的复杂度是O(1),所以长度位N的bitset的一次单点操作是O(1),整体操作是O(N/w),其中w=32.(long long 是64). 然后Bits…