CF732D Exams 题解
CF732D Exams
题目描述
Vasiliy has an exam period which will continue for \(n\) days. He has to pass exams on \(m\) subjects. Subjects are numbered from 1 to \(m\) .
About every day we know exam for which one of \(m\) subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number \(a_{i}\) — the number of days he should prepare to pass the exam number \(i\) . Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during \(a_{i}\) days for the exam number \(i\) . He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
输入格式
The first line contains two integers $ n$ and \(m\) ( \(1<=n,m<=10^{5}\) ) — the number of days in the exam period and the number of subjects.
The second line contains nn integers \(d_{1},d_{2},...,d_{n}\) ( \(0<=d_{i}<=m\) ), where \(d_{i}\) is the number of subject, the exam of which can be passed on the day number \(i\) . If \(d_{i}\) equals 0, it is not allowed to pass any exams on the day number \(i\) .
The third line contains \(m\) positive integers \(a_{1},a_{2},...,a_{m}\) ( \(1<=a_{i}<=10^{5}\) ), where \(a_{i}\) is the number of days that are needed to prepare before passing the exam on the subject \(i\) .
输出格式
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
题意翻译
一个人有m门科目需要考试，每一门科目需要a[i]（1<=i<=m）的复习时间(复习时间可以不用连续)，并且有一份n天的考试安排表，其中d[i]表示第i天能考第i门科目，（1<=i<=n）假如d[i]为0，就代表这一天没有任何科目的考试。试求这个人最少在第几天顺利通过所有考试？（注：这个人一天要么只能考试，要么就只能复习）
输入输出样例
输入 #1
7 2
0 1 0 2 1 0 2
2 1
输出 #1
5
输入 #2
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
输出 #2
9
输入 #3
5 1
1 1 1 1 1
5
输出 #3
-1
说明/提示
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it.
【思路】
二分答案 +贪心
【题目大意】
每一天你可以完成任务或者积攒能量
每一个任务的完成都需要消耗一定的能量
每一天都只能完成一个特定的任务或者不能完成任务
求最早什么时候完成全部的任务
【题目分析】
从题意中可以看出如果能够在第i天完成全部的任务
那么一定能够在第i+1天完成全部的任务
所以用二分答案就很显然了吧
【核心思路】
二分完成需要的天数
不过这里怎么判断某个天数到底行不行呢？
因为只需要判断行不行而不需要判断优不优
所以只需要知道能不能完成就可以了
不管完成的过程优不优
那就考虑最差的能够完成的情况
就是在规定时间内
每一个任务最后一次出现的时候能够把它完成掉就是可以的
所以在不是最后一次出现的情况下就囤积能量
在最后一次出现的地方就完成
如果完成不了那就是不行
到最后再检查一下一共m个任务是不是每一个任务都完成了
【完整代码】
#include<iostream>
#include<cstdio>
#include<cstring>
#define int long long
using namespace std;
int read()
{
int sum = 0,fg = 1;
char c = getchar();
while(c < '0' || c > '9')
{
if(c == '-')fg = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
{
sum = sum * 10 + c - '0';
c = getchar();
}
return sum * fg;
}
const int Max = 100005;
int d[Max];
int a[Max];
int t[Max];
int n,m;
bool check(int x)
{
memset(t,0,sizeof(t));
int tot = 0;
int acioi = 0;
for(register int i = 1;i <= x;++ i)
t[d[i]] ++;
for(register int i = 1;i <= x;i ++)
{
if(d[i] != 0)
{
t[d[i]] -= 1;
if(t[d[i]] == 0)
{
acioi ++;
tot -= a[d[i]];
if(tot < 0)
return false;
}
else
tot ++;
}
else
tot ++;
}
for(register int i = 1;i <= m;++ i)
if(t[i] != 0)
return false;
if(acioi == m)
return true;
else
return false;
}
signed main()
{
// freopen("generals.in","r",stdin);
// freopen("generals.out","w",stdout);
n = read(),m = read();
for(register int i = 1;i <= n;++ i)
d[i] = read();
for(register int i = 1;i <= m;++ i)
a[i] = read();
int l = 1,r = n + 1;
while(l < r)
{
int mid = (l + r) >> 1;
if(check(mid))r = mid;
else l = mid + 1;
}
if(l == n + 1)
{
cout << -1 << endl;
return 0;
}
cout << l << endl;
return 0;
}
