【BZOJ】3315: [Usaco2013 Nov]Pogo-Cow（dp）
http://www.lydsy.com/JudgeOnline/problem.php?id=3315
果然自己太弱。
想不出dp方程啊。。
其实，以后记住。。。与上一个状态或下一个状态有关，，可以开一维或多维。。
（这题暴力n^3都能a。。。。。。。。。。。。。。。。。
f(i, j)表示i个由j转移过来的最大价值
则
f(i, j)=max{f(j, k)}+p[i] 1<=k<j且x[i]-x[j]>=x[j]-x[k]
f(i, j)=max{f(i, j), f(j, 0)}
而这题还要注意！正反向都要dp！！！否则就要错。。QAQ
暴力：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr2(a, b, c) for1(i, 1, b) { for1(j, 1, c) cout << a[i][j]; cout << endl; }
#define printarr1(a, b) for1(i, 1, b) cout << a[i]; cout << endl
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
const int N=1005;
int f[N][N], n, ans;
struct dat { int x, w; }a[N];
bool cmp(const dat &a, const dat &b) { return a.x<b.x; }
int main() {
read(n);
for1(i, 1, n) read(a[i].x), read(a[i].w);
sort(a+1, a+1+n, cmp);
for1(i, 1, n) for1(j, 0, i-1) {
for1(k, 1, j-1) if(a[i].x-a[j].x>=a[j].x-a[k].x)
f[i][j]=max(f[i][j], f[j][k]);
f[i][j]=max(f[i][j], f[j][0]);
f[i][j]+=a[i].w;
ans=max(f[i][j], ans);
}
CC(f, 0);
for3(i, n, 1) for1(j, i+1, n+1) {
for1(k, j+1, n) if(a[i].x-a[j].x<=a[j].x-a[k].x)
f[i][j]=max(f[i][j], f[j][k]);
f[i][j]=max(f[i][j], f[j][n+1]);
f[i][j]+=a[i].w;
ans=max(f[i][j], ans);
}
print(ans);
return 0;
}
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
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
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。
