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【DP】 Codeforces Round #286 A - Mr. Kitayuta, the Treasure Hunter

2015-01-19 08:48 393 查看
注意到步数的改变量最多为250,然后就是简单DP了。。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100000
#define maxm 10005
#define eps 1e-10
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
//#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

int a[maxn];
int dp[maxn][505];
int n, d;

void read(void)
{
int x;
scanf("%d%d", &n, &d);
for(int i = 1; i <= n; i++) scanf("%d", &x), a[x]++;
}

void work(void)
{
dp[d][250] = 1 + a[d];
int ans = 0;
for(int i = 1; i <= 30000; i++) {
for(int j = 0; j <= 500; j++) {
if(!dp[i][j]) continue;
int t = j - 250 + d;
ans = max(ans, dp[i][j]);
if(t == 1) {
dp[i+t][j] = max(dp[i+t][j], dp[i][j] + a[i+t]);
dp[i+t+1][j+1] = max(dp[i+t+1][j+1], dp[i][j] + a[i+t+1]);
}
else {
dp[i+t-1][j-1] = max(dp[i+t-1][j-1], dp[i][j] + a[i+t-1]);
dp[i+t][j] = max(dp[i+t][j], dp[i][j] + a[i+t]);
dp[i+t+1][j+1] = max(dp[i+t+1][j+1], dp[i][j] + a[i+t+1]);
}
}
}
printf("%d\n", --ans);
}

int main(void)
{
read();
work();

return 0;
}
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