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); }