以前学斜率优化觉得好难
现在莫名觉得简单 （雾）
(fee[i])维护的一个前缀和
(sum[i])维护的成品的前缀和
(dis[i])维护(0->i)的距离
易得状态转移方程
[dp[i] = min(dp[i],dp[j] + fee[i] - fee[j] - sum[j] * (dis[i] - dis[j]) + cost[i]);
]
时间复杂度 (O(n^2)) (n_{max}=1e6) (TlE)
考虑优化
方程中有乘积
考虑斜率优化
易得
[frac{dp[j] - fee[j] + sum[j] * dis[j] - (dp[k] - fee[k] + sum[k] * dis[k])}{(sum[j] - sum[k])}<dis[i]
]
(Code)
#include <set>
#include <stack>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define reg register int
#define isdigit(x) ('0' <= x&&x <= '9')
template<typename T>
inline T Read(T Type)
{
T x = 0,f = 1;
char a = getchar();
while(!isdigit(a)) {if(a == '-') f = -1;a = getchar();}
while(isdigit(a)) {x = (x << 1) + (x << 3) + (a ^ '0');a = getchar();}
return x * f;
}
const int MAXN = 1e6 + 10;
int q[MAXN];
typedef long long ll;
ll sum[MAXN],dp[MAXN],dis[MAXN],fee[MAXN],cost[MAXN];
#define M(x) (x) * (x)
inline ll Get(int j,int k)
{
return dp[j] - fee[j] + sum[j] * dis[j] - (dp[k] - fee[k] + sum[k] * dis[k]);
}
inline ll Get2(int j,int k)
{
return sum[j] - sum[k];
}
inline ll Getdp(int i,int j)
{
return dp[j] + fee[i] - fee[j] - sum[j] * (dis[i] - dis[j]) + cost[i];
}
int main()
{
// dp[j] - fee[j] - sum[j] * (dis[i] - dis[j]) <
// dp[k] - fee[k] - sum[k] * (dis[i] - dis[k])
// dp[j] - fee[j] + sum[j] * dis[j] - (dp[k] - fee[k] + sum[k] * dis[k]) < (sum[j] - sum[k]) * dis[i]
int n = Read(1);
for(reg i = 1;i <= n;i++)
{
int x = Read(1),p = Read(1),c = Read(1);
dis[i] = x,sum[i] = sum[i - 1] + p,cost[i] = c;
fee[i] = fee[i - 1] + (dis[i] - dis[i - 1]) * sum[i - 1];
}
// memset(dp,0x3f3f3f,sizeof(dp));
// dp[0] = 0;
// for(reg i = 1;i <= n;i++)
// {
// for(reg j = 0;j < i;j++)
// {
// dp[i] = min(dp[i],dp[j] + fee[i] - fee[j] - sum[j] * (dis[i] - dis[j]) + cost[i]);
// }
// }
// printf("%lld
",dp[n]);
int head = 0,tail = 1;
for(reg i = 1;i <= n;i++)
{
while(head + 1 < tail&&1.0 * Get(q[head + 1],q[head]) / Get2(q[head + 1],q[head]) <= dis[i])
head++;
dp[i] = Getdp(i,q[head]);
while(head + 1 < tail&&1.0 * Get(i,q[tail - 1]) / Get2(i,q[tail - 1]) <= 1.0 * Get(q[tail - 1],q[tail - 2]) / Get2(q[tail - 1],q[tail - 2]))
tail--;
q[tail] = i;
tail++;
}
printf("%lld
",dp[n]);
return 0;
}
(Solution) (II)
考虑决策单调性
证明略