Codeforces Round #416 (Div. 2) C Vladik and Memorable Trip
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根据题目要求,目的地相同的人必须在一个车厢中,或者坐下趟车。因此,每个对人有两种决策:(1)上车(2)不上车
上车就必须保证,所有该目的的人之间的乘客也必须上车,这样中间的人同理,这个区间会不断扩张,直到不会有区间外的人跟这个区间里的某人目的地相同为止。剩下的区间又会变成子问题,记忆化求解即可。
解题代码如下:
#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#include<map>#include<string>#include<queue>#include<vector>#include<list>#include<bitset>//#pragma comment(linker,"/STACK:1024000000,1024000000")using namespace std;typedef long long ll;#define INF 0x3f3f3f3fint n;int a[5005];int dp[5005],fir[5005],lst[5005];void read(){ cin>>n; for(int i=1;i<=n;i++) cin>>a[i];}void init(){ memset(fir,-1,sizeof fir); memset(lst,-1,sizeof lst); memset(dp,-1,sizeof dp); for(int i=1;i<=n;i++) { if(fir[a[i]]==-1) fir[a[i]]=i; lst[a[i]]=i; }}int get(int p){ if(p==0) return 0; if(dp[p]!=-1) return dp[p]; int res=0,l=fir[a[p]],i; for(i=p;i>=l;i--) { if(lst[a[i]]>p) break; l=min(l,fir[a[i]]); if(fir[a[i]]==i) res^=a[i]; } dp[p]=get(p-1); if(i<l) dp[p]=max(res+get(i),dp[p]); return dp[p];}void solve(){ printf("%d\n",get(n));}int main(){ read(); init(); solve(); return 0;}
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