题目链接
题意
将$n$本书的前若干本分成$k$段,使每一段的和的最大值最小。
思路
二分答案,然后动态规划check。接下来考虑怎样计算一个mid是否可行。
$dp[i]$表示以第$i$本书为结尾最多分几段,那么$dp[i] = max(dp[j]) + 1.$
在线段树维护这个前缀和就好了。1
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using namespace std;
typedef long long ll;
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
const int N = 2e5 + 5;
int seg[N << 2];
vector<ll> des;
int dp[N];
void push_up(int rt)
{
seg[rt] = dp[seg[lson]] < dp[seg[rson]]? seg[rson] : seg[lson];
}
ll sum[N];
void build(int L, int R, int rt)
{
if (L == R)
{
seg[rt] = 200001;
return;
}
int mid = (L + R) >> 1;
build(Lson);
build(Rson);
push_up(rt);
}
void update(int p, int v, int L, int R, int rt)
{
if (L == R)
{
if (dp[seg[rt]] < dp[p])
seg[rt] = p;
return;
}
int mid = (L + R) >> 1;
if (v <= mid)
update(p, v, Lson);
else
update(p, v, Rson);
push_up(rt);
}
int query(int l, int r, int L, int R, int rt)
{
if (l > r)
return 200001;
if (l <= L && r >= R)
{
return seg[rt];
}
int mid = (L + R) >> 1;
int ans = 200001;
if (l <= mid)
{
int t = query(l, r, Lson);
ans = dp[ans] < dp[t]? t : ans;
}
if (r > mid)
{
int t = query(l, r, Rson);
ans = dp[ans] < dp[t]? t : ans;
}
return ans;
}
int get_id(ll v)
{
return lower_bound(des.begin(), des.end(), v) - des.begin() + 1;
}
int k, n;
bool check(ll mid)
{
//dbg(mid);
build(1, des.size(), 1);
dp[0] = 0;
dp[200001] = -1;
update(0, get_id(0), 1, des.size(), 1);
for (int i = 1; i <= n; i++)
{
ll s = sum[i] - mid;
int v = get_id(s);
//dbg(s, v);
int t = query(v, des.size(), 1, des.size(), 1);
if (dp[t] == -1)
dp[i] = -1;
else
dp[i] = dp[t] + 1;
// dbg(i, dp[i], t, dp[t]);
update(i, get_id(sum[i]), 1, des.size(), 1);
}
for (int i = 1; i <= n; i++)
if (dp[i] >= k)
return true;
return false;
}
int a[N];
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
scanf("%d%d", &n, &k);
des.clear();
sum[0] = 0;
des.push_back(0);
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
//des.push_back(a[i] * 1ll);
sum[i] = sum[i - 1] + a[i];
des.push_back(sum[i]);
}
sort(des.begin(), des.end());
des.erase(unique(des.begin(), des.end()), des.end());
ll l = -200000000000000ll, r = 200000000000000ll;
// ll l = -2, r = 4;
while (l < r)
{
ll mid = (l + r) >> 1;
if (check(mid))
r = mid;
else
l = mid + 1;
}
printf("%lld\n", l);
}
return 0;
}