[Usaco2017 Open]Modern Art 2

Description

Having become bored with standard 2-dimensional artwork (and also frustrated at others copying her work), the great bovine artist Picowso has decided to switch to a more minimalist, 1-dimensional style.Although, her paintings can now be described by a 1-dimensional array of colors of length N(1≤N≤100,000), her painting style remains unchanged: she starts with a blank canvas and layers upon it a sequence of "rectangles" of paint, which in this 1-dimensional case are simply intervals. Sheuses each of the colors 1…Nexactly once, although just as before, some colors might end up being completely covered up by the end.To Picowso's great dismay, her competitor Moonet seems to have figured out how to copy even these 1-dimensional paintings, using a similar strategy to the preceding problem: Moonet will paint a set ofdisjoint intervals, wait for them to dry, then paint another set of disjoint intervals, and so on.Moonet can only paint at most one interval of each color over the entire process. Please compute thenumber of such rounds needed for Moonet to copy a given 1-dimensional Picowso painting.

伟大的牛艺术家皮科沃已经厌倦了标准的二维艺术作品，也在伤心其他人复制她的作品，她决定转向更简约，一维的风格。尽管如此,她的作品现在可以表示描述颜色的一维数组长度N(1≤N≤100000)，她的绘画风格没有改变:她以一个空白的画布开始，一次涂色只能涂上连续几个单位的颜料，同样新的颜料可以完全覆盖旧的颜料，每次涂完要等上1day才能完全干，只有旧颜料干了以后才能用新颜料覆盖。皮科沃十分沮丧的是，她的对手Moonet似乎已经找到了如何复制这些一维的绘画，使用类似的策略。前面的问题：Moonet会画几组不相交区间，等待他们干，然后画几组不相交区间，等等。Moonet同一种颜色只能使用一次。请计算该轮Moonet复制一个给定的一维picowso绘画所需要的时间。

Input

The first line of input contains N, and the next N lines contain an integer in the range 0…Nindicating the color of each cell in the 1-dimensional painting (0 for a blank cell).

第一行为N，画条长度从第2行至N行每行一个数表示要涂颜色

Output

Please output the minimum number of rounds needed to copy this painting, or -1 if this could not have possibly been an authentic work of Picowso (i.e., if she could not have painted it using a layered sequence of intervals, one of each color).

输出一个整数表示最少天数。数据若不合法则输出-1

Sample Input

7

0

1

4

5

1

3

3

Sample Output

2

In this example, the interval of color 1 must be painted in an earlier round than the intervals of colors 4 and 5, so at least two rounds are needed.

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由于每次都是涂一段连续的区间，因此我们可以将其转换成括号序列，只要求出最大深度即可。不过这样写有个问题，假定我出现了一个1 3 1 3的序列，这显然是不可行的，所以括号序列上要记录一个信息，或者可以用栈来进行维护

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline int read(){
    int x=0,f=1;char ch=getchar();
    for (;ch<'0'||ch>'9';ch=getchar())    if (ch=='-')    f=-1;
    for (;ch>='0'&&ch<='9';ch=getchar())  x=(x<<1)+(x<<3)+ch-'0';
    return x*f;
}
inline void print(int x){
    if (x>=10)     print(x/10);
    putchar(x%10+'0');
}
const int N=1e5;
int stack[N+10],fir[N+10],val[N+10],las[N+10];
int main(){
    int n=read(),top=0,ans=0;
    for (int i=1,x;i<=n;i++){
        x=val[i]=read();
        if (!fir[x])    fir[x]=i;
	    las[x]=max(las[x],i);      //对于每种颜色，记录开始点和最后的结束点
    }
    val[las[0]=n+1]=0;
    for (int i=0;i<=n+1;i++){
        int x=val[i];
        if (i==fir[x])  stack[++top]=x,ans=max(ans,top);     //遇到开始点就入栈
        if (stack[top]!=x){puts("-1");return 0;}    //如果一个点不是不是开始点，并且不与栈顶元素相同，则不合法
        if (i==las[x])  top--;    //是结束点就减栈
    }
    printf("%d\n",ans-1);
    return 0;
}