2017 Multi-University Training Contest - Team 2&&hdu 6047 Maximum Sequence

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1450    Accepted Submission(s): 673

 

Sample Input

4

8 11 8 5

3 1 4 2

Sample Output

27

Hint

For the first sample: 1. Choose 2 from {bi}, then a_2…a_4 are available for a_5, and you can let a_5=a_2-2=9; 2. Choose 1 from {bi}, then a_1…a_5 are available for a_6, and you can let a_6=a_2-2=9;

Source

2017 Multi-University Training Contest - Team 2

题意：给你一个a数组  代表这个数组里的数值  给你一个b数组  代表a数组的位序  在b数组中选一个值  进行奇葩（解释不清楚）操作

8 11 8 5

3 1 4 2  选2   a变成-》8 11 8 5 9  在选1-》 8 11 8 5 9 9  再选3变成8 11 8 5 9 9 5  在选4-》8 11 8 5 9 9 5 4

求增加的数字和最大为多少

题解：其实将b数组排个序 从小到大选取 用一个优先队列维护一下是不是可以取到最大值  因为b数组的值不确定

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string.h>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<cmath>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const double PI=acos(-1.0);
const double eps=0.0000000001;
const int N=500000+10;
const ll mod=1e9+7;
int a[N],b[N];
struct node{
    int num;
    int pos;
    friend bool operator<(node aa,node bb){
        return aa.num<bb.num;
    }
};
priority_queue<node>q;
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
       // memset(c,0,sizeof(c));
        while(!q.empty())q.pop();
        node c;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            c.num=a[i]-i;
            c.pos=i;
            q.push(c);
        }
        //cout<<q.top().num<<endl;
        for(int i=1;i<=n;i++)scanf("%d",&b[i]);
        sort(b+1,b+1+n);
        ll ans=0;
        node t;
        int tt=n+1;
        for(int i=1;i<=n;i++){
            while(b[i]>q.top().pos)q.pop();
            if(q.empty()==1)break;
            c=q.top();
            ans=(ans+c.num)%mod;
            t.num=c.num-tt;
            t.pos=tt;
            q.push(t);
            tt++;
            //cout<<c.num<<endl;
        }
        ans%=mod;
        printf("%I64d\n",ans);
    }
}