Dilthey's Blog

引用来自http://www.cnblogs.com/iiyiyi/p/4738644.html的思路【题目大意】给出C头奶牛的SAT成绩和申请奖学金，选出N头牛，使得总奖学金在≤F的情况下奶牛SAT成绩的中位数最大。【思路】假设before[i]表示前i头奶牛中n/2头奶牛奖学金总额的最小值，而after[i]表示后i头奶牛中n/2头奶牛奖学金总额的最小值。将C头奶牛按照SAT

把牛按minSPF从小到大排序一下，防晒霜按SPF排序一下。然后遍历防晒霜lontion[1...L]，对于lotion[i]，把所有满足minSPF小于等于lontion[i].SPF的牛都找出来，入队。我们将这个队列定义为按cow[i].maxSPF的从小到大排序的优先队列，那么，每次取出的队首，都是当前要求最苛刻（SPF要求范围最小的）那头牛，先满足这些牛，如果当前这 种防晒霜能满足

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 23060    Accepted Submission(s): 9172Problem DescriptionIn the new year party, e

Time limit: 3.000 second限时：3.000秒 Background背景Many areas of Computer Science use simple, abstract domains for both analytical and empirical studies. For example, an early AI study of p

题目思路请参考：http://blog.youkuaiyun.com/acdreamers/article/details/8692384RMQ请参考：http://blog.youkuaiyun.com/liang5630/article/details/7917702代码请参考：#include#include#include#include#include#define MAXN 1003u

RMQ应用题。解题思路参考：http://blog.youkuaiyun.com/libin56842/article/details/46482803#include#include#include#include#define MAXN 100000+5using namespace std;int num[MAXN],a[MAXN];int n,q,seg_num

#include#include#define lowbit(x) x&(-x)typedef long long ll;using namespace std;ll n,q,num,root;string s;int main(){ scanf("%I64d%I64d",&n,&q); root=(n+1)/2; for(ll q_i=1;q_i<=q;q_i++) {

抽象来说就是求逆序对的问题，虽然放在树状数组的章节里，但不一定要用树状数组，归并排序也可以……#include#include#define MAX 1000*1000+3using namespace std;struct Road{ int u,v;}road[MAX],tmp[MAX];int n,m,k;long long cnt;bool cmp(Road a,Ro

#include#define MAX_N 20000+5int n,a[MAX_N],c[100000+5],MAX_ai;int pre_more[MAX_N],pre_less[MAX_N],suf_more[MAX_N],suf_less[MAX_N];long long ans;int lowbit(int x){return x&(-x);}void add(int i,i