本文深入探讨了AI音视频处理领域中的视频分割与语义识别技术，介绍了视频分割的基本概念、常见算法及其实现方式，同时详细阐述了语义识别在实际应用中的关键步骤和挑战，旨在为读者提供全面的技术视角。
Alena has successfully passed the entrance exams to the university and is now looking forward to start studying.
One two-hour lesson at the Russian university is traditionally called a pair, it lasts for two academic hours (an academic hour is equal to 45 minutes).
The University works in such a way that every day it holds exactly n lessons. Depending on the schedule of a particular group of students, on a given day, some pairs may actually contain classes, but some may be empty (such pairs are called breaks).
The official website of the university has already published the schedule for tomorrow for Alena's group. Thus, for each of the n pairs she knows if there will be a class at that time or not.
Alena's House is far from the university, so if there are breaks, she doesn't always go home. Alena has time to go home only if the break consists of at least two free pairs in a row, otherwise she waits for the next pair at the university.
Of course, Alena does not want to be sleepy during pairs, so she will sleep as long as possible, and will only come to the first pair that is presented in her schedule. Similarly, if there are no more pairs, then Alena immediately goes home.
Alena appreciates the time spent at home, so she always goes home when it is possible, and returns to the university only at the beginning of the next pair. Help Alena determine for how many pairs she will stay at the university. Note that during some pairs Alena may be at the university waiting for the upcoming pair.
The first line of the input contains a positive integer n (1 ≤ n ≤ 100) — the number of lessons at the university.
The second line contains n numbers ai (0 ≤ ai ≤ 1). Number ai equals 0, if Alena doesn't have the i-th pairs, otherwise it is equal to 1. Numbers a1, a2, ..., an are separated by spaces.
Print a single number — the number of pairs during which Alena stays at the university.
5 0 1 0 1 1
4
7 1 0 1 0 0 1 0
4
1 0
0
In the first sample Alena stays at the university from the second to the fifth pair, inclusive, during the third pair she will be it the university waiting for the next pair.
In the last sample Alena doesn't have a single pair, so she spends all the time at home.
题意：一天有n个课时，但是不是每个课时都有课，(1表示该课时有课，0表示该课时没课)，因为Alena家离学校比较远，所以如果这节有课直到下一节有课之间只有小于等于一个【没课的】课时，那么Alena就留在学校等待下一节课，否则就回家，求Alena在学校的时间【每个课时算1】
思路：题意相当于：两个1之间如果少于等于一个0时间，则继续留在学校【累加】，如果两个1之间有2个0或以上那么就回家【跳过】
#include<iostream>
using namespace std;
const int MAXN=105;
int a[MAXN];
int main()
{
int n,total=0;
cin>>n;
for(int i=0;i<n;i++)
{
cin>>a[i];
}
for(int i=0;i<n;i++)
{
if(a[i])
{
total++;
}
else
{
if((i-1>=0)&&(i+1<n)&&(a[i-1]&&a[i+1]))
{
total++;
}
}
}
cout<<total<<endl;
return 0;
}
A little boy Laurenty has been playing his favourite game Nota for quite a while and is now very hungry. The boy wants to make sausage and cheese sandwiches, but first, he needs to buy a sausage and some cheese.
The town where Laurenty lives in is not large. The houses in it are located in two rows, n houses in each row. Laurenty lives in the very last house of the second row. The only shop in town is placed in the first house of the first row.
The first and second rows are separated with the main avenue of the city. The adjacent houses of one row are separated by streets.
Each crosswalk of a street or an avenue has some traffic lights. In order to cross the street, you need to press a button on the traffic light, wait for a while for the green light and cross the street. Different traffic lights can have different waiting time.
The traffic light on the crosswalk from the j-th house of the i-th row to the (j + 1)-th house of the same row has waiting time equal to aij (1 ≤ i ≤ 2, 1 ≤ j ≤ n - 1). For the traffic light on the crossing from the j-th house of one row to the j-th house of another row the waiting time equals bj (1 ≤ j ≤ n). The city doesn't have any other crossings.
The boy wants to get to the store, buy the products and go back. The main avenue of the city is wide enough, so the boy wants to cross it exactly once on the way to the store and exactly once on the way back home. The boy would get bored if he had to walk the same way again, so he wants the way home to be different from the way to the store in at least one crossing.
Figure to the first
sample.Help Laurenty determine the minimum total time he needs to wait at the crossroads.
The first line of the input contains integer n (2 ≤ n ≤ 50) — the number of houses in each row.
Each of the next two lines contains n - 1 space-separated integer — values aij (1 ≤ aij ≤ 100).
The last line contains n space-separated integers bj (1 ≤ bj ≤ 100).
Print a single integer — the least total time Laurenty needs to wait at the crossroads, given that he crosses the avenue only once both on his way to the store and on his way back home.
4 1 2 3 3 2 1 3 2 2 3
12
3 1 2 3 3 2 1 3
11
2 1 1 1 1
4
The first sample is shown on the figure above.
In the second sample, Laurenty's path can look as follows:
- Laurenty crosses the avenue, the waiting time is 3;
- Laurenty uses the second crossing in the first row, the waiting time is 2;
- Laurenty uses the first crossing in the first row, the waiting time is 1;
- Laurenty uses the first crossing in the first row, the waiting time is 1;
- Laurenty crosses the avenue, the waiting time is 1;
- Laurenty uses the second crossing in the second row, the waiting time is 3.
In the last sample Laurenty visits all the crossings, so the answer is 4.
题意：就和例图的一样，从下面的n个位置去到上面的第一个位置，然后再从上面的第一个位置回到下面的
第n个位置，要求在来回两次的路径不能一样，并且中间的过道只能走一次，求最短的来回值。
思路：如果中间的过道没限制可能会难一点，但是因为中间的过道只能走一遍，所以就暴力找从哪个位置走过道就行了，
然后回来再从n-1个过道里面再选一个就行。
#include<iostream>
using namespace std;
const int MAXN=55;
#define INF 65535
int up[MAXN],down[MAXN],street[MAXN];
int main()
{
int n,a,index,min_go=INF,min_break=INF;
cin>>n;
up[n-1]=down[0]=0;
for(int i=n-2;i>=0;i--)
{
cin>>up[i];
}
for(int i=n-2;i>=0;i--)
{
up[i]+=up[i+1];
}
for(int i=n-1;i>=1;i--)
{
cin>>down[i];
}
for(int i=1;i<=n-1;i++)
{
down[i]+=down[i-1];
}
for(int i=n-1;i>=0;i--)
{
cin>>street[i];
}
for(int i=0;i<n;i++)
{
if(up[i]+street[i]+down[i]<min_go)
{
min_go=up[i]+street[i]+down[i];
index=i;
}
}
for(int i=0;i<n;i++)
{
if(i==index)
{
continue;
}
if(up[i]+street[i]+down[i]<min_break)
{
min_break=up[i]+street[i]+down[i];
}
}
cout<<min_break+min_go<<endl;
return 0;
}p:每个孩子的初始血量，d：当血量小于0时，则该孩子对他后面的孩子造成d点伤害，并且血量小于0时，该孩子就会跑
问牙医能帮几个孩子看牙，分别是哪个孩子
思路：主要问题在于题目的理解，模拟一下即可。
有几个坑点：①.因为数据可能到10^6，如果第一个孩子进去之后后面n-1个血量都小于0的话，【如果n个人的d都是极限的10^的话，那么会爆int，所以
要用long long 】 ②如果造成的伤害达到之前，该位置的孩子已经跑了的话，那么伤害就不会跳过他，对后面孩子排队的孩子造成伤害。如果造成伤害
后血量小于0，则还会对他后面的孩子造成d的伤害。
#include<iostream>
using namespace std;
const int MAXN = 4005;
long long int v[MAXN], p[MAXN], d[MAXN];
int main()
{
int n, ans = 0;
int inter[MAXN];
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> v[i] >> d[i] >> p[i];
}
for (int i = 1; i <= n; i++)
{
if (p[i]<0)//该孩子已经跑路
{
continue;
}
inter[ans++] = i;//该孩子进入门里，看医生
long long int tempv = v[i], tempd=0;
for (int j = i + 1; j <= n; j++)
{
if (p[j] < 0)//孩子已经跑路
{
continue;
}
if (p[j] >= 0)//孩子还在排队
{
p[j] -= (tempv + tempd);
tempv = (tempv - 1) > 0 ? (tempv - 1) : 0;
if (p[j] < 0)
{
tempd += d[j];
}
}
}
}
cout << ans << endl;
for (int i = 0; i<ans; i++)
{
cout << inter[i];
if (i != ans - 1)
{
cout << ' ';
}
}
cout << endl;
//system("pause");
return 0;
}
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