2019ICPC南昌網絡賽 I題 Max answer 暴力+路徑壓縮

題意：給出5e5個數，每個數大小範圍-1e5到1e5， 求一個最大的區間權值 = （區間中每個數的和）乘（區間中最小的數）

思路：暴力+路徑壓縮，枚舉每個點爲區間中最小的數，每個數有一個l和r代表區間左端點和右端點

代碼：

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxx = 5e5+5;
long long a[maxx];
int l[maxx], r[maxx];
long long sum[maxx];
int main()
{
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
        l[i] = r[i] = i, scanf("%lld", &a[i]), sum[i] = sum[i-1] + a[i];
    a[0] = a[n+1] = -1e18;
    for(int i = 1; i <= n; ++i)
    {
        if(a[i] > 0)
            while(a[i] <= a[l[i]-1])
                l[i] = l[l[i]-1];
        else if(a[i] < 0)
        {
            long long mx = a[i];
            int resl = l[i];
            while(a[i] <= a[l[i]-1])
            {
                l[i] = l[l[i]-1];
                if(sum[i]-sum[l[i]-1] < mx)
                {
                    mx = sum[i]-sum[l[i]-1];
                    resl = l[i];
                }
            }
            l[i] = resl;
        }
    }
    for(int i = n; i >= 1; --i)
    {
        if(a[i] > 0)
            while(a[i] <= a[r[i]+1])
                r[i] = r[r[i]+1];
        else if(a[i] < 0)
        {
            long long mx = a[i];
            int resr = i;
            while(a[i] <= a[r[i]+1])
            {
                r[i] = r[r[i]+1];
                if(sum[r[i]]-sum[i-1] < mx)
                {
                    mx = sum[r[i]]-sum[i-1];
                    resr = r[i];
                }
            }
            r[i] = resr;
        }
    }
    long long pp = 0;
    for(int i = 1; i <= n; ++i)
        pp = max(pp, a[i]*(sum[r[i]]-sum[l[i]-1]));
    cout<<pp<<endl;
    return 0;
}
// 5 1 -1 -2 -2 -3