BZOJ 3315: [Usaco2013 Nov]Pogo-Cow( dp )

我真想吐槽USACO的数据弱..= = O(n^3)都能A....上面一个是O(n²), 一个是O(n^3)

O(n^3)做法, 先排序, dp(i, j) = max{ dp(j, p) } + w( i ) ( t <= p <= j ) 表示跳到第 i 个点, 上一个点是在 j 的最大得分, 其中t是满足条件的最小p.

我们在计算dp(i, j) (1 <= j <= i )时会发现, 随着 j 的递减, t也在不断地减小, 这样我们只要在dp过程中维护h(i, j)表示 max{ dp(i, x) } ( j <= x <= i ), 然后逆序枚举 j, 维护t即可. 时间复杂度O(n²)

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cmath>

#define rep(i, n) for(int i = 0; i < n; i++)

#define clr(x, c) memset(x, c, sizeof(x))

using namespace std;

const int maxn = 1009;

struct R {

int p, v;

inline void Read() {

scanf("%d%d", &p, &v);

}

} A[maxn];

bool cmpL(const R &a, const R &b) {

return a.p < b.p;

}

bool cmpR(const R &a, const R &b) {

return a.p > b.p;

}

int dp[maxn][maxn], h[maxn][maxn], n, ans = 0;

void work() {

clr(dp, 0), clr(h, 0);

h[0][0] = dp[0][0] = A[0].v;

for(int i = 1; i < n; ++i) {

int p = i - 1;

h[i][i] = dp[i][i] = A[i].v;

for(int j = i - 1; j >= 0; j--) {

while(p && (abs(A[i].p - A[j].p) >= abs(A[j].p - A[p - 1].p) || p > j)) p--;

if(abs(A[i].p - A[j].p) >= abs(A[j].p - A[p].p))

dp[i][j] = max(h[j][p], dp[i][j]);

ans = max(ans, dp[i][j] += A[i].v);

h[i][j] = max(h[i][j + 1], dp[i][j]);

}

int main() {

freopen("test.in", "r", stdin);

cin >> n;

rep(i, n) A[i].Read();

sort(A, A + n, cmpL);

work();

sort(A, A + n, cmpR);

work();

cout << ans << "\n";

return 0;

}

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3315: [Usaco2013 Nov]Pogo-Cow

Time Limit: 3 Sec Memory Limit: 128 MB
Submit: 185 Solved: 100
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Description

In an ill-conceived attempt to enhance the mobility of his prize cow Bessie, Farmer John has attached a pogo stick to each of Bessie's legs. Bessie can now hop around quickly throughout the farm, but she has not yet learned how to slow down. To help train Bessie to hop with greater control, Farmer John sets up a practice course for her along a straight one-dimensional path across his farm. At various distinct positions on the path, he places N targets on which Bessie should try to land (1 <= N <= 1000). Target i is located at position x(i), and is worth p(i) points if Bessie lands on it. Bessie starts at the location of any target of her choosing and is allowed to move in only one direction, hopping from target to target. Each hop must cover at least as much distance as the previous hop, and must land on a target. Bessie receives credit for every target she touches (including the initial target on which she starts). Please compute the maximum number of points she can obtain.

一个坐标轴有N个点，每跳到一个点会获得该点的分数，并只能朝同一个方向跳，但是每一次的跳跃的距离必须不小于前一次的跳跃距离，起始点任选，求能获得的最大分数。

Input

* Line 1: The integer N.

* Lines 2..1+N: Line i+1 contains x(i) and p(i), each an integer in the range 0..1,000,000.

Output

* Line 1: The maximum number of points Bessie can receive.

Sample Input

6
5 6
1 1
10 5
7 6
4 8
8 10

INPUT DETAILS: There are 6 targets. The first is at position x=5 and is worth 6 points, and so on.

Sample Output

25
OUTPUT DETAILS: Bessie hops from position x=4 (8 points) to position x=5 (6 points) to position x=7 (6 points) to position x=10 (5 points).

从坐标为4的点，跳到坐标为5的，再到坐标为7和，再到坐标为10的。

HINT

Source

Silver

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