【codeforces】Far Relative’s Problem
2016-07-22 20:05
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Far Relative’s ProblemTime Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d& %I64uSubmit StatusDescriptionFamil Door wants to celebrate his birthday with his friends from Far Far Away. He has n friends and each of them can come to the party in a specific range of days of the year from ai to bi.Of course, Famil Door wants to have as many friends celebrating together with him as possible.Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party.Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends thatmay present at the party.InputThe first line of the input contains a single integer n (1 ≤ n ≤ 5000) — then number of Famil Door's friends.Then follow n lines, that describe the friends. Each line starts with a capital letter 'F' for female friends and with a capital letter 'M'for male friends. Then follow two integers ai and bi (1 ≤ ai ≤ bi ≤ 366),providing that the i-th friend can come to the party from day ai to day bi inclusive.OutputPrint the maximum number of people that may come to Famil Door's party.Sample InputInput
4 M 151 307 F 343 352 F 117 145 M 24 128Output
2Input
6 M 128 130 F 128 131 F 131 140 F 131 141 M 131 200 M 140 200Output
4HintIn the first sample, friends 3 and 4 can come on any day in range [117, 128].In the second sample, friends with indices 3, 4, 5 and 6 cancome on day 140.解题说明:此题是找到重复最多的区间数目,同时保证男女人数要一样。比较简单的方法是穷举,在男女每个区间上都进行标记,然从头遍历,找到一个最大值。由于比数据范围比较少,所以可以直接记录每个点出现的次数
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#include<stdio.h> #include<math.h> #include<algorithm> #include<iostream> using namespace std; int f[400],m[400],n,t=0; char c; int main() { int a,b; cin >> n; for(int i=0;i<n;i++) { cin>>c>>a>>b; if(c=='M') { for(int i=a;i<=b;i++) m[i]++; } else { for(int i=a;i<=b;i++) f[i]++; } } for(int i=0;i<=400;i++) { if(t<2*min(m[i],f[i])) t=2*min(m[i],f[i]); } cout<<t<<endl; return 0; }
把#include<iostream> cin cout换成scanf printf不知道为啥就出错了。。
int F[400],M[400],N,rs = 0; char c;放在main函数里面也会运行出错!!Don‘t know why
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