http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1672
题意:
思路:
其实这就是一个经典的区间贪心问题,只需要按照左端点排序,然后用优先队列维护,每次将右端点最小的点出队列。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pll;
const int INF = 0x3f3f3f3f;
const int maxn = +; int n,k,m;
int a[maxn];
ll sum[maxn]; priority_queue< int,vector<int>,greater<int> > q; struct node
{
int left, right;
bool operator< (const node& rhs) const
{
return left<rhs.left||(left==rhs.left && right<rhs.right);
}
}p[maxn]; int main()
{
//freopen("in.txt","r",stdin);
while(~scanf("%d%d%d",&n,&k,&m))
{
sum[]=;
for(int i=;i<=n;i++)
{
scanf("%d",&a[i]);
sum[i]=sum[i-]+a[i];
}
for(int i=;i<m;i++)
scanf("%d%d",&p[i].left,&p[i].right);
sort(p,p+m);
while(!q.empty()) q.pop();
ll ans=;
for(int i=;i<m;i++)
{
q.push(p[i].right);
if(q.size()==k)
{
ans=max(ans,sum[q.top()]-sum[p[i].left-]);
q.pop();
}
}
printf("%lld\n",ans);
}
return ;
}