Codeforces Round #169 (Div. 2) C. Little Girl and Maximum Sum（线段树区间更新）

C. Little Girl and Maximum Sum

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

The little girl loves the problems on array queries very much.

One day she came across a rather well-known problem: you've got an array of n elements (the elements of the array are indexed starting from 1); also, there are q queries, each one is defined by a pair of integers li, ri (1 ≤ li ≤ ri ≤ n). You need to find for each query the sum of elements of the array with indexes from li to ri, inclusive.

The little girl found the problem rather boring. She decided to reorder the array elements before replying to the queries in a way that makes the sum of query replies maximum possible. Your task is to find the value of this maximum sum.

Input
The first line contains two space-separated integers n (1 ≤ n ≤ 2·105) and q (1 ≤ q ≤ 2·105) — the number of elements in the array and the number of queries, correspondingly.

The next line contains n space-separated integers ai (1 ≤ ai ≤ 2·105) — the array elements.

Each of the following q lines contains two space-separated integers li and ri (1 ≤ li ≤ ri ≤ n) — the i-th query.

Output
In a single line print a single integer — the maximum sum of query replies after the array elements are reordered.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Sample test(s)

Input

3 3
5 3 2
1 2
2 3
1 3

Output

25

Input

5 3
5 2 4 1 3
1 5
2 3
2 3

Output

33

对每个数a[i]，统计其被访问的次数cnt[i],将a和cnt分别从小到大排列，maxSum = a[0]*cnt[0]+a[1]*cnt[1]+……

#include <iostream>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
#define cti const int
#define dg(i) cout << "*" << i << endl;

cti MAXN = 200002;
struct Node
{
int r, l;
LL inc;  //新增查询次数
LL cnt;  //该区间被查询次数
}tree[MAXN<<2|1];
int lastNode;  //记录最后一个结点的下标
LL a[MAXN];
bool tag[MAXN<<2|1];  //标记是否叶子
LL cnt[MAXN]; //记录叶子的访问次数

bool cmp (const LL& x, const LL& y) {return x > y;}

void Build(int left, int right, int rt)
{
if(lastNode < rt) lastNode = rt;
tree[rt].l = left;
tree[rt].r = right;
if(left == right)
{
tag[rt] = true;
return ;
}
int mid = (left + right) >> 1;
Build(left, mid, rt << 1);
Build(mid + 1, right, rt << 1 | 1);
}

void Update(int from, int to, int rt)
{
if(from <= tree[rt].l && tree[rt].r <= to)
{
tree[rt].inc++;
tree[rt].cnt++;
return ;
}
if(tree[rt].inc)
{
tree[rt<<1].cnt += tree[rt].inc;
tree[rt<<1|1].cnt += tree[rt].inc;
tree[rt<<1].inc += tree[rt].inc;
tree[rt<<1|1].inc += tree[rt].inc;
tree[rt].inc = 0;
}
int mid = (tree[rt].l + tree[rt].r) >> 1;
if(from <= mid) Update(from, to, rt << 1);
if(to > mid) Update(from, to, rt << 1 | 1);
}

int main()
{
int n, q, from, to;
while(scanf("%d %d", &n, &q) != EOF)
{
memset(tag, false, sizeof(tag));
for(int i = 0; i < n; i++) scanf("%I64d", &a[i]);
for(int i = 1; i < (n<<2); i++)
tree[i].cnt = tree[i].inc = 0;
lastNode = 0;
Build(1, n, 1);
while(q--)
{
scanf("%d %d", &from, &to);
Update(from, to, 1);
}
//此循环保证区间更新彻底往下传递
for(int i = 1, j = 0; i <= lastNode; i++)
{
if(tree[i].inc && (i<<1) < lastNode)
{
tree[i<<1].cnt += tree[i].inc;
tree[i<<1|1].cnt += tree[i].inc;
tree[i<<1].inc += tree[i].inc;
tree[i<<1|1].inc += tree[i].inc;
}
if(tag[i])  //叶子
{
cnt[j++] = tree[i].cnt;
}
}
sort(a, a + n, cmp);
sort(cnt, cnt + n, cmp);
LL sum = 0;
for(int i = 0; i < n; i++)
sum += (cnt[i] * a[i]);
printf("%I64d\n", sum);
}
return 0;
}