cf599C Day at the Beach（区间合并）

C. Day at the Beach
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
One day Squidward, Spongebob and Patrick decided to go to the beach. Unfortunately, the weather was bad, so the friends were unable to ride waves. However, they decided to spent their time building sand castles.

At the end of the day there were n castles built by friends. Castles are numbered from 1 to n, and the height of the i-th castle is equal to hi. When friends were about to leave, Squidward noticed, that castles are not ordered by their height, and this looks ugly. Now friends are going to reorder the castles in a way to obtain that condition hi ≤ hi + 1 holds for all i from 1 to n - 1.

Squidward suggested the following process of sorting castles:

Castles are split into blocks — groups of consecutive castles. Therefore the block from i to j will include castles i, i + 1, …, j. A block may consist of a single castle.
The partitioning is chosen in such a way that every castle is a part of exactly one block.
Each block is sorted independently from other blocks, that is the sequence hi, hi + 1, …, hj becomes sorted.
The partitioning should satisfy the condition that after each block is sorted, the sequence hi becomes sorted too. This may always be achieved by saying that the whole sequence is a single block.
Even Patrick understands that increasing the number of blocks in partitioning will ease the sorting process. Now friends ask you to count the maximum possible number of blocks in a partitioning that satisfies all the above requirements.

Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of castles Spongebob, Patrick and Squidward made from sand during the day.

The next line contains n integers hi (1 ≤ hi ≤ 109). The i-th of these integers corresponds to the height of the i-th castle.

Output
Print the maximum possible number of blocks in a valid partitioning.

Sample test(s)
input
3
1 2 3
output
3
input
4
2 1 3 2
output
2
Note
In the first sample the partitioning looks like that: [1][2][3].

In the second sample the partitioning is: [2, 1][3, 2]

就是给一个序列，然后划分区间，每个区间自己排序之后的结果和整体排序的结果是一样的，问最多划分多少区间。

当时按照区间的思想想了一下，每个数排序之后都是有位置的，那么这个数现在的位置到排序后的位置就是这个数的最短区间。
然后猜想一下（因为有的数会有多个）
找出这个数的上下界up,down
1.如果现在位置在上下界内，区间就是[i,i];
2.如果在下界以下，就是[i,down];
3.上界以上，就是[up,i];
然后得到n个区间，接着区间合并，看有多少个独立区间。

当时没有完整证明出来，现在还是想不通，只能从yy角度上想通一点，找了半天也没有反例，而且终测也过了，看样应该有道理，再想想吧。

#include<cstdio>#include<cstring>#include<iostream>#include<queue>#include<vector>#include<algorithm>#include<string>#include<cmath>#include<set>#include<map>#include<vector>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f;const int maxn = 1005;int a[100005],b[100005];struct Node{    int l,r;}node[100005];bool cmp(Node a,Node b){    if(a.l == b.l)return a.r < b.r;    return a.l < b.l;}int main(){    #ifdef LOCAL    freopen("C:\\Users\\ΡΡ\\Desktop\\in.txt","r",stdin);    //freopen("C:\\Users\\ΡΡ\\Desktop\\out.txt","w",stdout);    #endif // LOCAL    int n;    scanf("%d",&n);    for(int i = 1;i <= n;i++)        scanf("%d",&a[i]);    for(int i = 1;i <= n;i++)b[i] = a[i];    sort(b + 1,b + 1 + n);    for(int i = 1;i <= n;i++)    {        int down = lower_bound(b + 1,b + 1 + n,a[i]) - b;        int up = upper_bound(b + 1,b + 1 + n,a[i]) - b - 1;        if(down <= i &&i <= up)node[i].l = node[i].r = i;        else if(i < down)        {            node[i].l = i;node[i].r = down;        }        else        {            node[i].r = i;node[i].l = up;        }    }    /*for(int i = 1;i <= n;i++)        cout<<node[i].l<<" "<<node[i].r<<endl;*/    sort(node + 1,node + 1 + n,cmp);    int ans,l,r;    ans = 1;l = node[1].l;r = node[1].r;    for(int i = 2;i <= n;i++)    {        if(node[i].l > r)        {            ans++;l = node[i].l;r = node[i].r;        }        else            r = max(r,node[i].r);    }    printf("%d\n",ans);    return 0;}