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洛谷 P3195 [HNOI2008] 玩具装箱
题目介绍
链接：https://www.luogu.com.cn/prob…
解题报告
解法一（TLE）
看到题首先写出暴力版本 dp
#include <iostream>
#include <algorithm>
typedef long long ll;
#define read(x) scanf("%lld", &x)
using namespace std;
const int N = 50010;
ll sumc[N], f[N], n, L;
int main() {read(n);
read(L);
ll c;
for (int i = 1; i <= n; i++) {read(c);
sumc[i] = sumc[i - 1] + c;
}
for (int j = 1; j <= n; j++) {f[j] = 0x3f3f3f3f3f3f3f3f;
for (int i = 0; i < j; i++) {f[j] = min(f[j], f[i] + (j - i - 1 + sumc[j] - sumc[i] - L) * (j - i - 1 + sumc[j] - sumc[i] - L));
}
}
cout << f[n];
return 0;
}易知上述解法是 O(n^2) 的工夫复杂度，而该题样例给到了 5 乘 10 的 4 次方，这个解法是会 TLE 的。因而应该是能够依据动静转移方程优化的。
上述解法提交后和意料的一样，有三个测试点 TLE 了，然而另外七个测试点 AC 证实动静转移方程没有问题，因而接下来能够抛开题意分心依据该动静转移方程进行优化
解法二（AC）依据斜率优化
动静转移方程为
f[j] = min(f[j], f[i] + (j - i - 1 + sumc[j] - sumc[i] - L) * (j - i - 1 + sumc[j] - sumc[i] - L));令
$$
a[i]=sumc[i]+i
$$
$$
b[i]=sumc[i]+i+L+1
$$
简化计算
上式可转化为
$$
f[j] = f[i] + (a[j] – b[i]) ^ 2
$$
开展得
$$
f[i]+b[i]^2=2a[j]*b[i]-a[j]^2+f[j]
$$
当 j 确定时，a[j] 为常量并且大于 0，将 f[i]+b[i]^2 作为 y，b[i] 作为 x
则可简化为
$$
y=2a[j]*x-a[j]^2+f[j]
$$
2a[j] > 0，所以斜率大于 0
当 y 轴截距最小时，f[j] 最小
将斜率为 2a[j] 的线向上移，第一个碰到的点即为能使 f[j] 最小的点，凸包原理得点 i 处斜率是第一个大于绿线斜率的点
依据凸包原理，只需保护一个凸包即可
这题 a[j] 也是枯燥递增的，因而后面不合乎的点必定不会再被用到了，能够间接抛弃，只需保护无限几个点即可
代码如下：
#include <iostream>
#include <algorithm>
typedef long long ll;
#define read(x) scanf("%lld", &x)
using namespace std;
const int N = 50010;
ll sumc[N], f[N], n, L;
ll hh, tt, q[N];
ll A(ll i) {return sumc[i] + i;
}
ll B(ll i) {return (ll) sumc[i] + i + L + 1;
}
ll Y(ll i) {return f[i] + B(i) * B(i);
}
ll X(ll i) {return B(i);
}
double slope(ll i1, ll i2) {return (double) (Y(i2) - Y(i1)) / (double) ((X(i2) - X(i1)));
}
int main() {read(n);
read(L);
ll c;
hh = tt = 0;
for (int i = 1; i <= n; i++) {read(c);
sumc[i] = sumc[i - 1] + c;
}
for (int j = 1; j <= n; j++) {while (hh < tt && Y(q[hh + 1]) - Y(q[hh]) < (X(q[hh + 1]) - X(q[hh])) * 2 * A(j)) {hh++;}
f[j] = f[q[hh]] + (A(j) - B(q[hh])) * (A(j) - B(q[hh]));
while (hh < tt && slope(j, q[tt - 1]) <= slope(q[tt], q[tt - 1])) {tt--;}
q[++tt] = j;
}
cout << f[n];
return 0;
}
胜利 AC
P4072 [SDOI2016] 征途
题目介绍
链接：https://www.luogu.com.cn/prob…
解题报告
解法一
思考动静转移方程
f[i] [j] 示意走了 i 段当初在 j 地
上一步为 f[i – 1] [k] k 可能为 0~j- 1 地
写出以下代码
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
typedef long long ll;
#define read(x) scanf("%d", &x)
using namespace std;
const int N = 3010;
int n, m, sum[N];
ll f[N][N];
int main() {read(n);
read(m);
for (int i = 1; i <= n; i++) {
int c;
read(c);
sum[i] = sum[i - 1] + c;
}
memset(f, 0x3f, sizeof f);
f[0][0] = 0;
for (int i = 1; i <= m; i++) {for (int j = 1; n - j >= m - i; j++) {for (int k = 0; k < j; k++) {f[i][j] = min(f[i][j], f[i - 1][k] + (sum[j] - sum[k]) * (sum[j] - sum[k]));
}
}
}
ll res = f[m][n] * m - sum[n] * sum[n];
cout << res;
return 0;
}
提交后可看到因为复杂度较高，有一个样例仍然是过不了的
解法二
将动静转移方程移项得：
$$
f[i-1][k] + sum[k] ^2 = f[i,j] + 2* sum[j] * sum[k] – sum[j] ^2
$$
令 f[i-1] [k] + sum[k] ^2 = y
sum[k] = x
斜率为 2 *sum[j] > 0
因为 sum[j] 为定量，所以使截距最小即可
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
typedef long long ll;
#define read(x) scanf("%d", &x)
using namespace std;
const int N = 3010;
int n, m, sum[N], q[N], hh, tt;
ll f[N][N];
ll X(int k) {return sum[k];
}
ll Y(int i, int k) {return f[i - 1][k] + sum[k] * sum[k];
}
double slope(int i, int k1, int k2) {return (double) (Y(i, k1) - Y(i, k2)) / (double) (X(k1) - X(k2));
}
int main() {read(n);
read(m);
for (int i = 1; i <= n; i++) {
int c;
read(c);
sum[i] = sum[i - 1] + c;
}
memset(f, 0x3f, sizeof f);
for (int i = 1; i <= n; i++) {f[1][i] = sum[i] * sum[i];
}
f[0][0] = 0;
for (int i = 2; i <= m; i++) {
hh = tt = 0;
for (int j = 1; n - j >= m - i; j++) {while (hh < tt && slope(i, q[hh], q[hh + 1]) < 2 * sum[j]) {hh++;}
f[i][j] = f[i - 1][q[hh]] + (sum[j] - sum[q[hh]]) * (sum[j] - sum[q[hh]]);
while (hh < tt && slope(i, j, q[tt - 1]) <= slope(i, q[tt], q[tt - 1])) {tt--;}
q[++tt] = j;
}
}
ll res = f[m][n] * m - sum[n] * sum[n];
cout << res;
return 0;
}胜利 AC
AcWing Q301 工作安顿
题目介绍
链接：https://www.acwing.com/proble…
解题报告
解法一
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
typedef long long ll;
#define read(x) scanf("%d", &x)
#define read(x, y) scanf("%d%d", &x, &y)
using namespace std;
const int N = 5010;
int n, s, sumt[N], sumc[N], f[N];
int main() {read(n, s);
for (int i = 1; i <= n; i++) {
int t, c;
read(t, c);
sumt[i] = sumt[i - 1] + t;
sumc[i] = sumc[i - 1] + c;
}
for (int i = 1; i <= n; i++) {f[i] = 0x3f3f3f3f;
for (int j = 0; j < i; j++) {f[i] = min(f[i], f[j] + (sumc[n] - sumc[j]) * s + (sumc[i] - sumc[j]) * sumt[i]);
}
}
cout << f[n];
return 0;
}同上优化
解法二
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
typedef long long ll;
#define read(x, y) scanf("%lld%lld", &x, &y)
using namespace std;
const int N = 300010;
ll n, s, sumt[N], sumc[N], f[N];
int hh, tt, q[N];
ll X(int i) {return sumc[i];
}
ll Y(int i) {return f[i];
}
double slope(int i1, int i2) {return (double) (Y(i1) - Y(i2)) / (double) (X(i1) - X(i2));
}
int main() {read(n, s);
for (int i = 1; i <= n; i++) {
ll t, c;
read(t, c);
sumt[i] = sumt[i - 1] + t;
sumc[i] = sumc[i - 1] + c;
}
for (int i = 1; i <= n; i++) {while (hh < tt && slope(q[hh], q[hh + 1]) < s + sumt[i]) {hh++;}
f[i] = f[q[hh]] + (sumc[n] - sumc[q[hh]]) * s + (sumc[i] - sumc[q[hh]]) * sumt[i];
while (hh < tt && slope(i, q[tt - 1]) <= slope(q[tt], q[tt - 1])) {tt--;}
q[++tt] = i;
}
cout << f[n];
return 0;
}
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