南昌网络赛 Max answer（单调栈）

题意：给你n个数，让你找到数组中区间的最大价值，区间价值定义为区间的最小值*区间的数字加和

思路：用单调栈处理出左右l，r.然后用线段树或者st维护区间的最值，如果a[i]<0就左侧min，右侧max，大于0就反一下，（为什么比赛的时候一直不过啊）

代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

const int maxn = 5e5 + 7;
int stk[maxn], top;
int a[maxn];
int n;
int l[maxn], r[maxn];
LL pre[maxn];
LL minn[maxn << 2], maxe[maxn << 2];

void build(int i , int l, int r)
{
    if(l == r) {
        maxe[i] = minn[i] = pre[l];
        return ;
    }
    int mid = (l + r) >> 1;
    build(i << 1, l, mid);
    build(i << 1 | 1, mid + 1, r);
    minn[i] = min(minn[i << 1], minn[i << 1 | 1]);
    maxe[i] = max(maxe[i << 1], maxe[i << 1 | 1]);
}

LL query1(int i, int L, int R,int l, int r)
{
    if(l <= L && R <= r) {
        return minn[i];
    }
    int mid = (L + R) >> 1;
    LL res = 0x3f3f3f3f3f3f3f3fLL;
   if(l <= mid) res = min(res, query1(i << 1, L, mid, l, r));
   if(mid < r) res = min(res, query1(i << 1 | 1, mid + 1, R, l, r));
   return res;
}
LL query2(int i, int L, int R, int l, int r)
{
    if(l <= L && R <= r) {
        return maxe[i];
    }
    int mid = (L + R) >> 1;
    LL res = -0x3f3f3f3f3f3f3f3fLL;
    if(l <= mid) res = max(res, query2(i << 1, L, mid, l, r));
    if(mid < r) res = max(res, query2(i << 1 | 1, mid + 1, R, l, r));
    return res;
}
int main()
{
  while(~scanf("%d", &n)) {
    top = 0; pre[0] = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      pre[i] = pre[i - 1] + a[i];
    }
    top = 0;
    for (int i = 1; i <= n; i++) {
      while(a[i] <= a[stk[top]] && top) top--;
      if(!top) l[i] = 0;
      else l[i] = stk[top];
      stk[++top] = i;
    }
    top = 0;
    for (int i = n; i; i--) {
      while(a[i] <= a[stk[top]] && top) top--;
      if(!top) r[i] = n;
      else r[i] = stk[top] - 1;
      stk[++top] = i;
    }
    build(1, 0, n);
    LL ans = 0;
    for (int i = 1; i <= n; i++) {
      LL res = 1LL * a[i];
      if(a[i] < 0) res *= query1(1, 0, n, i, r[i]) - query2(1, 0, n, l[i], i - 1);
      else res *= query2(1, 0, n, i, r[i]) - query1(1, 0, n, l[i], i - 1);
      // printf("test %d %d res == %lld %lld %lld l == %d r == %d\n", i, a[i], res, query2(1, 0, n, i, r[i]), query1(1, 0, n, l[i], i - 1), l[i], r[i]);
      ans = max(ans, res);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
/*
1
-5
*/