【2017多校】HDU6047 Maximum Sequence 【贪心】

显然每个bi都要被选择一次,每次贪心的选出an+k=max{aj | bi<=j<n+k}
求出ai的后缀最大值a′i=max(a[j] | j>=i)
对bi从小到大排序,依次选取即可,顺便维护一个maxVal=max(a[i] | i>n)

#include<stdio.h>#include<iostream>#include<stdlib.h>#include<algorithm>#include<vector>#include<string.h>#include<string>#include<math.h>#include<memory.h>#include<queue>#define ll long long#define pii pair<int,int>#define pll pair<ll,ll>#define MEM(a,x) memset(a,x,sizeof(a))#define lowbit(x) ((x)&-(x))using namespace std;//const int inf=0x3f3f3f3f;const int MOD = 1e9+7;const int N = 250000 + 5;const int inf=MOD;int a[N],b[N];ll slove(int n){    ll maxVal=-inf;    ll ans=0;    for(int i=1;i<=n;++i){        int idx=b[i];        ll t=max(maxVal,(ll)a[idx]);        ans=(ans+t)%MOD;        maxVal=max(maxVal,(ll)a[idx]-(n+i));    }    return ans;}int main(){    //freopen("/home/lu/code/r.txt","r",stdin);    //freopen("/home/lu/code/w.txt","w",stdout);    int n;    while(~scanf("%d",&n)){        for(int i=1;i<=n;++i){            scanf("%d",&a[i]);            a[i]-=i;        }        for(int i=n-1;i>=1;--i){            a[i]=max(a[i],a[i+1]);        }        for(int i=1;i<=n;++i){            scanf("%d",&b[i]);        }        sort(b+1,b+n+1);        printf("%lld\n",slove(n));    }    return 0;}