Evolution Game
题目描述
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?
输入
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).
输出
For each test case, display one line containing the maximum possible number of changes.
样例输入
5 5
5 3 2 1 4
样例输出
4
题意
角从小变大，在眼睛范围w内的进化，问最多能进化几次？
题解
一个比较明显的dp，先对角进行排序，从后向前遍历，满足在w范围内的就dp+1
代码
#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
",x)
#define pri2(x,y) printf("%d %d
",x,y)
#define pri3(x,y,z) printf("%d %d %d
",x,y,z)
#define prl(x) printf("%lld
",x)
#define prl2(x,y) printf("%lld %lld
",x,y)
#define prl3(x,y,z) printf("%lld %lld %lld
",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);
}