ICPC China Nanchang National Invitational - I. Max answer(线段树+ST)

题目链接
N个数字求一个区间使得∑i=li=ra[i]×∑i=li=rmin(a[i])\sum_{i=l}^{i=r} a[i] × \sum_{i=l}^{i=r}min(a[i])∑i=li=ra[i]×∑i=li=rmin(a[i])最大

思路

枚举区间的最小值为a[i]a[i]a[i]，根据ST表二分找到它最左 LLL 和最右 RRR端点，这是保证区间[L,R][L, R][L,R]的最小值为a[i]a[i]a[i]
preprepre记录前缀和
sufsufsuf记录后缀和

找到区间之后分两种情况

a[i]≥0a[i] \ge0a[i]≥0
Ans=a[i]×(MaxPre[i,R]−Pre[i−1]+MaxSuf[L,i]−Suf[i])Ans = a[i]×(MaxPre[i, R]-Pre[i-1] + MaxSuf[L,i]-Suf[i])Ans=a[i]×(MaxPre[i,R]−Pre[i−1]+MaxSuf[L,i]−Suf[i])
因为a[i]≥0a[i] \ge0a[i]≥0所以我们在区间[i,R][i, R][i,R]找到一个最大的前缀和，在区间[L,i][L, i][L,i]找到一个最大的后缀和这样就能保证值最大
a[i]<0a[i] <0a[i]<0
和上一种情况相反
Ans=a[i]×(MinPre[i,R]−Pre[i−1]+MinSuf[L,i]−Suf[i])Ans = a[i]×(MinPre[i, R]-Pre[i-1] + MinSuf[L,i]-Suf[i])Ans=a[i]×(MinPre[i,R]−Pre[i−1]+MinSuf[L,i]−Suf[i])
因为a[i]<0a[i] <0a[i]<0所以我们在区间[i,R][i, R][i,R]找到一个最小的前缀和，在区间[L,i][L, i][L,i]找到一个最小的后缀和这样就能保证值最大

Ac之路很坎坷，刚开始全用ST表维护最大值最小值，结果MLE（ST很快但是空间也很高），然后把求后缀的ST表用线段树写，还是MLE。前后缀都带用线段树写，两个线段树代码重复太高而且变量名也不好起，还没用C++写过类，学习下杰哥用类写的线段树。Orz。。。

AC

#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
using namespace std;
const int maxn = 5e5 + 5;int a[maxn];
int dp[maxn][20];
LL pre[maxn], suf[maxn];
void ST(int n) {for (int i = 1; i <= n; ++i) {dp[i][0] = a[i];}for (int j = 1; j <= log2(n); ++j) {for (int i = 1; i+(1<<j)-1 <= n; ++i) {dp[i][j] = min(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);}}
}
LL RMQ(int l, int r) {int len = r - l + 1;int x = log2(len);return min(dp[l][x], dp[r - (1<<x)+1][x]);
}class Seg{LL Max[maxn<<2], Min[maxn<<2];public: void build(int rt, int l, int r, LL a[]) {if (l == r) {Max[rt] = a[l];Min[rt] = a[l];return;}build(lc, l, mid, a);build(rc, mid+1, r, a);Max[rt] = max(Max[lc], Max[rc]);Min[rt] = min(Min[lc], Min[rc]);   }public: LL query_min(int rt, int l, int r, int ql, int qr) {if (r < ql || l > qr) return 1e18;if (l >= ql && r <= qr) return Min[rt];LL tmp1 = query_min(lc, l, mid, ql, qr);LL tmp2 = query_min(rc, mid+1, r, ql, qr);return min(tmp1, tmp2);}public: LL query_max(int rt, int l, int r, int ql, int qr) {if (r < ql || l > qr) return -1e18;if (l >= ql && r <= qr) return Max[rt];LL tmp1 = query_max(lc, l, mid, ql, qr);LL tmp2 = query_max(rc, mid+1, r, ql, qr);return max(tmp1, tmp2);}
}Pre, Suf;int main () {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int n;scanf("%d", &n); for (int i = 1; i <= n; ++i) {scanf("%d", &a[i]);pre[i] = pre[i-1] + a[i];}for (int i = n; i >= 1; --i) {suf[i] = suf[i+1] + a[i];}ST(n);Pre.build(1, 1, n, pre);Suf.build(1, 1, n, suf);LL ans = -1e18;for (int i = 1; i <= n; ++i) {int left, right, l = i, r = n;while (l <= r) {if (RMQ(l, mid) < a[i]) r = mid - 1;else l = mid + 1;}right = r;l = 1, r = i;while (l <= r) {if (RMQ(mid, r) < a[i]) l = mid + 1;else r = mid - 1;}left = l;if (a[i] >= 0) {LL R = Pre.query_max(1, 1, n, i, right);LL L = Suf.query_max(1, 1, n, left, i);ans = max(ans, (R - pre[i-1] + L - suf[i]) * a[i]);}else {LL R = Pre.query_min(1, 1, n, i, right);LL L = Suf.query_min(1, 1, n, left, i);ans = max(ans, (R - pre[i-1] + L - suf[i]) * a[i]);}}cout << ans << endl;return 0;
}

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ICPC China Nanchang National Invitational - I. Max answer(线段树+ST)

思路

AC

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