【LG5017】[NOIP2018pj]摆渡车
题面
题解
震惊！普及组竟然考斜率优化？？？
当然有其他的方法
首先我们转化一下模型
此题可以变为：
在一根时间轴上有一些点，每个时间点\(i\)有一个权值\(c_i\)(即在\(i\)开始等待人数，没有则为\(0\))
要求选一些时间点，每个时间点间隔不小于\(m\)
使得每个点的权值乘上它与第一个大于等于它时间的已选择的时间点到它的距离之和最小 感觉讲得好复杂
设\(dp[i]\)表示当我们强制选时间点\(i\)的最小值
则有转移方程\(dp[i]=\min{dp[j]+\sum_{k=j+1}^i{(i-k)*c_k}}\) \((\)\(0\)\(\leq\)\(j\)\(\leq\)\(i-m\)\()\)
次数直接转移的复杂度为\(O(n^3)\)
考虑怎么优化，设
\(sum1[i]=\sum_{j=0}^i{c_j}\)
\(sum2[i]=\sum_{j=0}^i{c_j*j}\)
然后方程化为\(dp[i]=\min dp[j]+i*sum1[i]-i*sum1[j]-sum2[i]+sum2[j]\) \((\)\(0\)\(\leq\)\(j\)\(\leq\)\(i-m\)\()\)
此时复杂度为\(O(n^2)\)
继续优化，此时用上斜率优化
去掉\(min\)，则
\(dp[i]=dp[j]+i*sum1[i]-i*sum1[j]-sum2[i]+sum2[j]\)
移项得\(dp[j]+sum2[j]=i*sum1[j]+dp[i]-i*sum1[j]+sum2[i]\)
将\(dp[j]+sum2[j]\)视为\(y\)
将\(i\)视为\(k\)
将\(sum1[j]\)视为\(x\)
队列优化下凸壳即可
复杂度\(O(n)\)
但是好像会出现很多玄学问题啊~
下面这份使用\(double\)算斜率的代码交换\(slope\)函数中的\(i\)和\(j\)会\(WA\)
就是换成这样:
inline double slope(int j, int i) { return 1.0 * (y(j) - y(i)) / (x(i) == x(j) ? 1e-9 : x(j) - x(i)); } 但是好像理论上没问题啊？
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
using namespace std;
#define MAX_T 4100000
int N, M, c[MAX_T];
int maxt, sum1[MAX_T], sum2[MAX_T];
int dp[MAX_T], q[MAX_T];
#define y(i) (dp[i] + sum2[i])
#define k(i) (i)
#define x(i) (sum1[i])
inline double slope(int i, int j) { return 1.0 * (y(j) - y(i)) / (x(i) == x(j) ? 1e-9 : x(j) - x(i)); }
int main () {
cin >> N >> M;
for (int t, i = 1; i <= N; i++) cin >> t, c[t]++, maxt = max(maxt, t);
sum1[0] = c[0], sum2[0] = 0;
for (int i = 1; i < maxt + M; i++) {
sum1[i] = sum1[i - 1] + c[i];
sum2[i] = sum2[i - 1] + i * c[i];
}
int ans = INT_MAX;
int l = 1, r = 0;
for (int i = 0; i < maxt + M; i++) {
if (i - M >= 0) {
while (l < r && slope(q[r - 1], q[r]) >= slope(q[r], i - M)) --r;
q[++r] = i - M;
}
while (l < r && slope(q[l], q[l + 1]) <= k(i)) ++l;
dp[i] = i * sum1[i] - sum2[i];
int j = q[l]; if (l <= r) dp[i] = min(dp[i], dp[j] + i * sum1[i] - i * sum1[j] - sum2[i] + sum2[j]);
}
for (int i = maxt; i < maxt + M; i++) ans = min(ans, dp[i]);
cout << ans << endl;
return 0;
} 不用\(double\)不存在上面的问题:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
using namespace std;
#define MAX_T 4100000
int N, M, c[MAX_T];
int maxt, sum1[MAX_T], sum2[MAX_T];
int dp[MAX_T], q[MAX_T];
#define y(i) (dp[i] + sum2[i])
#define k(i) (i)
#define x(i) (sum1[i])
int main() {
cin >> N >> M;
for (int t, i = 1; i <= N; i++) cin >> t, c[t]++, maxt = max(maxt, t);
sum1[0] = c[0], sum2[0] = 0;
for (int i = 1; i < maxt + M; i++) {
sum1[i] = sum1[i - 1] + c[i];
sum2[i] = sum2[i - 1] + i * c[i];
}
int ans = INT_MAX;
int l = 1, r = 0;
for (int i = 0; i < maxt + M; i++) {
if (i - M >= 0) {
while (l < r && 1ll * (y(q[r]) - y(q[r - 1])) * (x(i - M) - x(q[r])) >=
1ll * (y(i - M) - y(q[r])) * (x(q[r]) - x(q[r - 1]))) -- r;
q[++r] = i - M;
}
while (l < r && 1ll * (y(q[l + 1]) - y(q[l])) <=
1ll * k(i) * (x(q[l + 1]) - x(q[l]))) ++l;
dp[i] = i * sum1[i] - sum2[i];
int j = q[l]; if (l <= r) dp[i] = min(dp[i], dp[j] + i * sum1[i] - i * sum1[j] - sum2[i] + sum2[j]);
}
for (int i = maxt; i < maxt + M; i++) ans = min(ans, dp[i]);
cout << ans << endl;
return 0;
}