Sereja and Two Sequences CodeForces - 425C (dp)

大意: 给定序列$a,b$, 每次可以任取两个相同大小的$a_i,b_j$删除$a_i,b_j$左侧所有元素, 花费为e, 得分1, 最后结束时必须再花费之前删除元素的个数, 不得分. 初始能量$s$, 求最大得分方案.

这题关键是注意到$\frac{s}{e}$的范围比较小, 直接暴力dp即可..

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#include <bitset>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

//head

#ifdef ONLINE_JUDGE

const int N = 1e6+10;

#else

const int N = 111;

#endif

int n, m, s, e;

vector<int> g[N];

int dp[N], a[N];

int main() {

	scanf("%d%d%d%d", &n, &m, &s, &e);

	REP(i,1,n) scanf("%d", a+i);

	REP(i,1,m) {

		int t;

		scanf("%d", &t);

		g[t].pb(i);

	}

	memset(dp,0x3f,sizeof dp);

	dp[0] = 0;

	int ans = 0;

	REP(i,1,n) PER(j,0,s/e) {

		auto t = upper_bound(g[a[i]].begin(),g[a[i]].end(),dp[j]);

		if (t==g[a[i]].end()) continue;

		dp[j+1] = min(dp[j+1], *t);

		if (dp[j+1]+i+e*(j+1)<=s) ans=max(ans,j+1);

	}

	printf("%d\n", ans);

}