Evolution Game

題目描述

In the fantasy world of ICPC there are magical beasts. As they grow, these beasts can change form, and every time they do they become more powerful. A beast cannot change form completely arbitrarily though. In each form a beast has n eyes and k horns, and these affect the changes it can make.

A beast can only change to a form with more horns than it currently has.
A beast can only change to a form that has a difference of at most w eyes. So, if the beast currently has n eyes it can change to a form with eyes in range [n - w, n + w].

A beast has one form for every number of eyes between 1 and N, and these forms will also have an associated number of horns. A beast can be born in any form. The question is, how powerful can one of these beasts become? In other words, how many times can a beast change form before it runs out of possibilities?

輸入

The first line contains two integers, N and w, that indicate, respectively, the maximum eye number, and the maximum eye difference allowed in a change (1 ≤ N ≤ 5000; 0 ≤ w ≤ N).
The next line contains N integers which represent the number of horns in each form. I.e. the ith number, h(i), is the number of horns the form with i eyes has (1 ≤ h(i) ≤ 1 000 000).

輸出

For each test case, display one line containing the maximum possible number of changes.

樣例輸入

5 5
5 3 2 1 4

樣例輸出

4

題意

角從小變大，在眼睛範圍w內的進化，問最多能進化幾次？

題解

一個比較明顯的dp，先對角進行排序，從後向前遍歷，滿足在w範圍內的就dp+1

代碼

#include<iostream>
#include<cstdio>  //EOF,NULL
#include<cstring> //memset
#include<cstdlib> //rand,srand,system,itoa(int),atoi(char[]),atof(),malloc
#include<limits.h> //INT_MAX
#include<bitset> // bitset<?> n
#include<cmath>  //ceil,floor,exp,log(e),log10(10),hypot(sqrt(x^2+y^2)),cbrt(sqrt(x^2+y^2+z^2))
#include<algorithm> //fill,reverse,next_permutation,__gcd,
#include<iomanip> //setw(set_min_width),setfill(char),setprecision(n),fixed,
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<utility>
#include<iterator>
#include<functional>
#include<map>
#include<set>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define scac(x) scanf("%c",&x)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define scl(x) scanf("%lld",&x)
#define scl2(x,y) scanf("%lld%lld",&x,&y)
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define pri(x) printf("%d\n",x)
#define pri2(x,y) printf("%d %d\n",x,y)
#define pri3(x,y,z) printf("%d %d %d\n",x,y,z)
#define prl(x) printf("%lld\n",x)
#define prl2(x,y) printf("%lld %lld\n",x,y)
#define prl3(x,y,z) printf("%lld %lld %lld\n",x,y,z)
#define ll long long
#define LL long long
#define read read()
#define pb push_back
#define mp make_pair
#define P pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(1.0)
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
#define N 5005
const int maxn = 2000005;
int n,w;

struct node
{
  int h,e,dp;
}a[N];
bool cmp(node a,node b)
{
  return a.h < b.h;
}
int main()
{
  sca2(n,w);
  rep(i,1,n+1)
  {
    sca(a[i].h);
    a[i].e = i;
  }
  sort(a+1,a+n+1,cmp);
  int l,r;
  for(int i = n; i >= 1;i--)
  {
    l = a[i].e - w,r = a[i].e + w;
    rep(j,i+1,n+1)
    {
      if(a[j].e >= l && a[j].e <= r && a[j].h > a[i].h)
      {
        a[i].dp = max(a[i].dp,a[j].dp+1);
      }
    }
  }
  int ans = a[1].dp;
  rep(i,2,n+1)
    ans = max(ans,a[i].dp);
  pri(ans);
}

upc組隊賽14 Evolution Game【dp】