Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

CF321E Ciel and Gondolas

LuoGu: CF321E Ciel and Gondolas

CF: E. Ciel and Gondolas

发现具有凸性,但是过不了。还需要四边形不等式优化。

要证明 $s _ {l, r + 1} + s _ {l + 1, r} \ge s _ {l, r} + s _ {l + 1, r + 1}$ 。使用数学分析法。
$$
\begin {array}{c}
& s _ {l, r + 1} + s _ {l + 1, r} \ge s _ {l, r} + s _ {l + 1, r + 1} \\
\Longrightarrow & s _ {l, r + 1} - s _ {l, r} \ge s _ {l + 1, r + 1} - s _ {l + 1, r} \\
\Longrightarrow & 2w _ {r + 1:l \to r} \ge 2 w _ {r + 1: l + 1 \to r}
\end {array}
$$
成立。

注意读入量巨大,要写快读。

查看代码
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#include <cstdio>
using namespace std;
const int N = 4e3 + 10, inf = 144e6;
int n, m, s[N][N], g[N][N];
struct Data
{
int t, s;
friend bool operator<(const Data &x, const Data &y)
{
return x.s < y.s || (x.s == y.s && x.t < y.t);
}
friend Data operator+(const Data &x, int k)
{
return (Data){x.t + 1, x.s + k};
}
} f[N];
Data cal(int k, int i)
{
return f[k] + g[k + 1][i];
}
struct Node
{
int p, l, r;
} q[N];
int hd, tl;
void check(int x)
{
hd = 1, tl = 0;
q[++tl] = (Node){0, 1, n};
for (int i = 1; i <= n; i++)
{
while (hd <= tl && q[hd].r < i)
hd++;
f[i] = cal(q[hd].p, i);
f[i].s += x;
int p = n + 1;
while (hd <= tl && cal(i, q[tl].l) < cal(q[tl].p, q[tl].l))
p = q[tl--].l;
if (hd <= tl && cal(i, q[tl].r) < cal(q[tl].p, q[tl].r))
{
int l = q[tl].l, r = q[tl].r;
while (l <= r)
{
int mid = l + r >> 1;
cal(i, mid) < cal(q[tl].p, mid) ? (p = mid, r = mid - 1) : l = mid + 1;
}
q[tl].r = p - 1;
}
if (p <= n)
q[++tl] = (Node){i, p, n};
}
}
void read(int &x)
{
x = 0;
char c = getchar();
while (c < '0' || c > '9')
c = getchar();
while (c >= '0' && c <= '9')
{
x = x * 10 + c - '0';
c = getchar();
}
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
{
read(s[i][j]);
s[i][j] += s[i][j - 1];
}
for (int k = 1; k <= n; k++)
for (int i = 1, j = k + 1; j <= n; i++, j++)
g[i][j] = g[i][j - 1] + s[j][j] - s[j][i - 1];
int l = 0, r = inf, res;
while (l <= r)
{
int mid = l + r >> 1;
check(mid);
f[n].t <= m ? (res = f[n].s - mid * m, r = mid - 1) : l = mid + 1;
}
printf("%d", res);
return 0;
}