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作者：@dwtfukgv
本文为作者原创，转载请注明出处：https://www.cnblogs.com/dwtfukgv/p/7246487.html

题意：给定一个序列，让你构造出一个序列，满足条件，且最大。条件是选取一个ai <= max{a[b[j], j]-j}

析：贪心，贪心策略就是先尽量产生大的，所以就是对于B序列尽量从头开始，由于数据比较大，采用桶排序，然后维护一个单调队列，使得最头上最大。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
 
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 250000 + 10;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r > 0 && r <= n && c > 0 && c <= m;
}
 
int num[maxn];
int a[maxn<<1];
int q[maxn<<1];
 
int main(){
  while(scanf("%d", &n) == 1){
    memset(num, 0, sizeof num);
    for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
    for(int i = 0; i < n; ++i){
      scanf("%d", &m);
      ++num[m];
    }
 
    int fro = 0, rear = 0;
    q[++rear] = 1;
    for(int i = 2; i <= n; ++i){
      while(rear > fro && a[i] - i >= a[q[rear]] - q[rear])  --rear;
      q[++rear] = i;
    }
    int cnt = 1;
    for(int i = n+1; i <= n+n; ++i){
      while(!num[cnt]) ++cnt;
      --num[cnt];
      while(q[fro+1] < cnt)  ++fro;
      a[i] = a[q[fro+1]] - q[fro+1];
      while(rear > fro && a[i] - i >= a[q[rear]] - q[rear])  --rear;
      q[++rear] = i;
    }
 
    int ans = 0;
    for(int i = n+1; i <= n+n; ++i)  ans = (ans + a[i]) % mod;
    printf("%d\n", ans);
  }
  return 0;
}