Codeforces Round #286 (Div. 2) C Mr. Kitayuta, the Treasure Hunter ( DP )

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题目链锁：　http://codeforces.com/contest/505/problem/C

题意：
有30001个岛，0~30000，给出n个岛上有宝藏，第一步跳跃距离为d，假设上一步跳跃距离为l，则当前可以跳l-1，l，l+1的距离

dp[i][j]表示在第i个点，上一步的跳跃距离l与第一步的跳跃距离相差为j，即 l = j + d
则
dp[i+l-1][j-1] = max ( dp[i+l-1][j-1], dp[i][j] + cnt[i+l-1] );
dp[i+l][j] = max( dp[l+i][j], dp[i][j] + cnt[i+l] );
dp[i+l+1][j+1] = max( dp[i+l+1][j+1], dp[i][j] + cnt[i+l+1] );
由于 ( d ) + ( d + 1 ) + ( d + 2 ) + ..... + ( d + 250 ) > 251d + 30000 所以取偏移量250；

#include<iostream>#include<cstring>#include<string>#include<cstdio>#include<stdio.h>#include<algorithm>#include<cmath>#include<set>#include<map>#include<queue>#include<vector>using namespace std;#define inf 0x3f3f3f3f#define eps 1e-9#define mod 1000000007#define FOR(i,s,t) for(int i = s; i < t; ++i )#define REP(i,s,t) for( int i = s; i <= t; ++i )#define LL long long#define ULL unsigned long long#define pii pair<int,int>#define MP make_pair#define lson id << 1 , l , m#define rson id << 1 | 1 , m + 1 , r #define maxn ( 30000+10 )#define maxe ( 200+10 )int dp[maxn][600], cnt[maxn];int main () {int n, d;while( cin>>n>>d ) {memset( cnt, 0, sizeof( cnt ) );for( int i = 0; i < n; ++i ) {int x;scanf("%d", &x );++cnt[x];}memset( dp, -1, sizeof( dp ) );dp[d][250] = cnt[d];int ans = cnt[d];for( int i = d; i <= 30000; ++i ) {for( int j = 0; j <= 500; ++j ) {if( dp[i][j] == -1 ) continue;int l = d + ( j - 250 );if( l > 0 && i + l <= 30000 ) {dp[i+l][j] = max( dp[l+i][j], dp[i][j] + cnt[i+l] );ans = max( ans, dp[i+l][j] );}if( l - 1 > 0 && i + l - 1 <= 30000 ) {dp[i+l-1][j-1] = max( dp[i+l-1][j-1] , dp[i][j] + cnt[i+l-1] );ans = max( ans, dp[i+l-1][j-1] );}if( l + 1 <= 30000 && l + 1 > 0 && i + l + 1 <= 30000 ) {dp[i+l+1][j+1] = max( dp[i+l+1][j+1], dp[i][j] + cnt[i+l+1] );ans = max( ans, dp[i+l+1][j+1] );}}}printf("%d\n", ans ); }}

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