洛谷P2120 [ZJOI2007]仓库建设 斜率优化DP
做的第一道斜率优化\(DP\)QwQ
原题链接1/原题链接2
首先考虑\(O(n^2)\)的做法：设\(f[i]\)表示在\(i\)处建仓库的最小费用，则有转移方程：
using namespace std;
define ll long long
define N 1000000
define re register
int n, X[N+5], P[N+5], C[N+5];
ll f[N+5];
inline char Getchar() {
static char buf[100000], *p1 = buf, p2 = buf;
return p1p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1p2)?EOF:p1++;
}
inline int read() {
int s = 0; re char c = Getchar();
while(c < '0' || c > '9') c = Getchar();
while(c >= '0' && c <= '9') s = s*10+c-'0', c = Getchar();
return s;
}
template
int main() {
n = read();
for(re int i = 1; i <= n; ++i) X[i] = read(), P[i] = read(), C[i] = read(), f[i] = 9223372036854775807LL;
f[0] = 0;
for(re int i = 1; i <= n; ++i) {
ll sum = 0, d = 0;
for(re int j = i-1; j >= 0; --j) {
f[i] = min(f[i], f[j]+sum+C[i]);
d += X[j+1]-X[j];
sum += P[j]*d;
}
}
printf("%lld", f[n]);
return 0;
}
那正解应该怎么做呢？经过观察，发现上面的转移方程已经无法再进行常规的优化了，此时我们需要斜率优化了。就拿这道题来说吧：
首先我们改写一下转移方程。若令：
<center>$sum1[i]=\sum\limits_{j=1}^{i}P[j]\ \ \ \ sum2[i]=\sum\limits_{j=1}^{i}P[j]X[j]$</center>
则$f[i]$可表示为：
<center>$f[i]=min\{f[j]+X[i](sum1[i]-sum1[j])-(sum2[i]-sum2[j])\}+C[i]$</center>
此时转移到了$i$，对于$j>k,j,k\in[0,i-1]$来说，若选$j$比选$k$更优，我们可以得到下式：
$$f[j]+X[i](sum1[i]-sum1[j])-(sum2[i]-sum2[j])<f[k]+X[i](sum1[i]-sum1[k])-(sum2[i]-sum2[k])\\
=>f[j]+sum2[j]-f[k]-sum2[k]<X[i]sum1[j]-X[i]sum1[k]\\
=>\frac{f[j]+sum2[j]-f[k]-sum2[k]}{sum1[j]-sum1[k]}<X[i]$$
当上式成立时，选$j$更优
此时我们令$y_i=f[i]+sum2[i],x_i=sum1[i]$，则上式表示的就是平面上两点$P_j$和$P_k$所确定的直线的斜率。如果我们把所有转移点的图像绘制出来的话，会长成下面这样：
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观察绿色的部分，我们会发现一些有趣的性质：
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分类讨论，有以下$3$种情况：
1.$X[i]>Slope(k,j)>Slope(j,i)$，$j$比$k$，$i$比$j$优
2.$Slope(k,j)>X[i]>Slope(j,i)$，$k$比$j$，$i$比$j$优
3.$Slope(k,j)>Slope(j,i)>X[i]$，$k$比$j$，$j$比$i$优
也就是说，在这种情况下，选$j$无论如何也不可能是最优的，于是我们可以把$j$去掉了。然后，得到的是一个下凸（是下凸吧）函数的图像：
<img src='data:img/jpg;base64,iVBORw0KGgoAAAANSUhEUgAAAzMAAAHMCAIAAABukmEEAAAAAXNSR0IArs4c6QAAAARnQU1BAACx
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此时我们再看看从这个图像上的哪一个点转移最优：假设$Slope(j,i)>X[i],Slope(k,j)<X[i]$，不难得出，$j$比左边所有的点都优，比右边的点也都优。所以我们从$j$点转移是最优的。又因为$X[i]$具有单调性，所以我们维护一个单调队列，每次从斜率大于等于$X[i]$的第一条直线的左端点转移就好了。
```cpp
#include <cstdio>
using namespace std;
#define N 1000000
#define ll long long
#define re register
int n, h = 1, t = 1, q[N+5], X[N+5], P, C[N+5];
ll f[N+5], sum1[N+5], sum2[N+5];
inline char Getchar(){
static char buf[2048], *p1=buf, *p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,2048,stdin),p1==p2)?EOF:*p1++;
}
inline int read() {
int s = 0; char c = Getchar();
while(c < '0' || c > '9') c = Getchar();
while(c >= '0' && c <= '9') s = s*10+c-'0', c = Getchar();
return s;
}
template <typename T> T min(T x, T y) { return x < y ? x : y; }
inline double getK(const int &j, const int &k) {
return 1.0*(f[j]+sum2[j]-f[k]-sum2[k])/(sum1[j]-sum1[k]);
}
int main() {
n = read();
for(int i = 1; i <= n; ++i) {
X[i] = read(), P = read(), C[i] = read();
sum1[i] = sum1[i-1]+P, sum2[i] = sum2[i-1]+P*X[i];
f[i] = X[i]*sum1[i]-sum2[i]+C[i];
}
for(int i = 1, j; i <= n; ++i) {
while(h+1 < t && getK(q[h], q[h+1]) < X[i]) h++;
j = q[h];
if(j) f[i] = min(f[i], f[j]+X[i]*(sum1[i]-sum1[j])-(sum2[i]-sum2[j])+C[i]);
while(h+1 < t && getK(q[t-2], q[t-1]) > getK(q[t-1], i)) t--;
q[t++] = i;
}
printf("%lld\n", f[n]);
return 0;
}
