題目連結

    題目大意可轉化為, 給出一系列區間[a, b], 將這些點按照區間重疊數目來排序.
    如果直接暴力, 每次回遍歷(b-a)次, 直接超時.
    a, b兩點正好是跳變點, 記錄下a, b點的狀態. 這個和電路中那些波的疊加有幾分像····

程式碼:

#include<cstdio>
#include<algorithm>
using namespace std;
int n, q, l, r;
int a[200001];
int b[200001];
long long ans;
int main()
{
    scanf("%d%d", &n, &q);
    for
 
 ( int i = 0; i < n; i ++ )
        scanf("%d", a + i);
    while ( q -- )
    {
        scanf("%d %d", &l, &r);
        b[l-1] ++;
        b[r]   --;
    }
    q = 0;

    for ( int i = 0; i < n; i ++ )
    {
        b[i] += q;
        q = b[i];
    }
    sort(a, a + n);
    sort(b, b + n);
    for
 
 ( int i = 0; i < n; i ++ )  ans += (long long)a[i]*b[i];
    printf("%I64d\n", ans);
    return 0;
}