luogu P2365 任务安排

嘟嘟嘟




如果常规dp，\(dp[i][j]\)表示前\(i\)个任务分\(j\)组，得到

\[dp[i][j] = min _ {k = 0} ^ {i - 1} (dp[k][j - 1] + (s * j + sumt[i]) * (sumc[i] - sumc[k]))
\]

复杂度是\(O(n ^ 3)\)的。

因此我们要换一个思路。

在执行一批任务时，我们虽然不知道之前机器启动过多少次，但是可以确定机器因执行这批人武而花费的启动时间为\(s\)，会累加到后面的任务上。

因此，令\(dp[i]\)表示把前\(i\)个任务分成若干批的最小费用，则

\[dp[i] = min_{j = 0} ^ {i - 1} (dp[j] + sumt[i] * (sumc[i] - sumc[j]) + s * (sumc[n] - sumc[j]))
\]

\(sumt[i] * (sumc[i] - sumc[j])\)表示的是不考虑机器启动时前\(i\)批任务的费用。之所以可以这么写，是因为后面的\(s * (sumc[n] - sumc[j])\)已经把他们的时间算进去了，即包含在了\(dp[j]\)中。

时间复杂度\(O(n ^ 2)\)。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 5e3 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, s;
int sumt[maxn], sumc[maxn];
int dp[maxn];

int main()
{
  n = read(); s = read();
  for(int i = 1, t, c; i <= n; ++i)
    {
      t = read(), sumt[i] = sumt[i - 1] + t;
      c = read(), sumc[i] = sumc[i - 1] + c;
    }
  Mem(dp, 0x3f); dp[0] = 0;
  for(int i = 1; i <= n; ++i)
    for(int j = 0; j < i; ++j)
      dp[i] = min(dp[i], dp[j] + sumt[i] * (sumc[i] - sumc[j]) + s * (sumc[n] - sumc[j]));
  write(dp[n]), enter;
  return 0;
}

上述算法已经能过此题，但还有一个$O(n)$的做——斜率优化。
简单来说就是对上述dp式进行变形，把常数、仅与$i$有关的项、仅与$j$有关的项以及$i, j$的乘积项分开。
具体维护下凸壳等想法不想讲了(懒)，以后填坑吧
先上代码
```c++
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e4 + 5;
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans = 10) write(x / 10);
putchar(x % 10 + '0');
}

int n, s;

ll sumt[maxn], sumc[maxn];

ll dp[maxn];

int q[maxn], l = 1, r = 1;

int main()

{

n = read(); s = read();

for(int i = 1, t, c; i <= n; ++i)

{

t = read(), sumt[i] = sumt[i - 1] + t;

c = read(), sumc[i] = sumc[i - 1] + c;

}

Mem(dp, 0x3f); dp[0] = 0;

for(int i = 1; i <= n; ++i)

{

while(l < r && (dp[q[l + 1]] - dp[q[l]]) <= (s + sumt[i]) * (sumc[q[l + 1]] - sumc[q[l]])) l++;

dp[i] = dp[q[l]] - (s + sumt[i]) * sumc[q[l]] + sumt[i] * sumc[i] + s * sumc[n];

while(l < r && (dp[q[r]] - dp[q[r - 1]]) * (sumc[i] - sumc[q[r]]) >= (dp[i] - dp[q[r]]) * (sumc[q[r]] - sumc[q[r - 1]])) r--;

q[++r] = i;

}

write(dp[n]), enter;

return 0;

}