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;}