Codeforces Round #332 (Div. 2)C. Day at the Beach
2016-08-23 11:04
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C. Day at the Beach
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard 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 tohi.
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.
Examples
input
output
input
output
Note
In the first sample the partitioning looks like that: [1][2][3].
In the second sample the partitioning is: [2, 1][3, 2]
题意:给出一个无序序列序列里面有n个数,可以将这个序列分成连续的k块将每块里的数字排序最终使得总序列有序问k的最大值。
/* ***********************************************
Author : ryc
Created Time : 2016-08-23 Tuesday
File Name : E:\acm\codeforces\332C.cpp
Language : c++
Copyright 2016 ryc All Rights Reserved
************************************************ */
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<list>
#include<vector>
#include<list>
#include<queue>
#include<map>
using namespace std;
typedef long long LL;
typedef pair<int,int>pii;
const int maxn=1000010;
int num[maxn];
struct Node{
int a,id;
}A[maxn];
bool cmp(Node a,Node b){
if(a.a==b.a)
return a.id<b.id;
return a.a<b.a;
}
int main()
{
int n;cin>>n;
for(int i=1;i<=n;++i){
scanf("%d",&A[i].a);
A[i].id=i;
}
sort(A+1,A+n+1,cmp);
int ans=0,Maxnum=0,cnt=0,l=0;
for(int i=1;i<=n;++i){
Maxnum=max(Maxnum,A[i].id);
if(A[i].id<=Maxnum){cnt++;}
if(cnt==Maxnum-l){
ans++;l=Maxnum;Maxnum=0;cnt=0;
}
}
printf("%d\n",ans);
return 0;
}
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard 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 tohi.
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.
Examples
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]
题意:给出一个无序序列序列里面有n个数,可以将这个序列分成连续的k块将每块里的数字排序最终使得总序列有序问k的最大值。
/* ***********************************************
Author : ryc
Created Time : 2016-08-23 Tuesday
File Name : E:\acm\codeforces\332C.cpp
Language : c++
Copyright 2016 ryc All Rights Reserved
************************************************ */
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<list>
#include<vector>
#include<list>
#include<queue>
#include<map>
using namespace std;
typedef long long LL;
typedef pair<int,int>pii;
const int maxn=1000010;
int num[maxn];
struct Node{
int a,id;
}A[maxn];
bool cmp(Node a,Node b){
if(a.a==b.a)
return a.id<b.id;
return a.a<b.a;
}
int main()
{
int n;cin>>n;
for(int i=1;i<=n;++i){
scanf("%d",&A[i].a);
A[i].id=i;
}
sort(A+1,A+n+1,cmp);
int ans=0,Maxnum=0,cnt=0,l=0;
for(int i=1;i<=n;++i){
Maxnum=max(Maxnum,A[i].id);
if(A[i].id<=Maxnum){cnt++;}
if(cnt==Maxnum-l){
ans++;l=Maxnum;Maxnum=0;cnt=0;
}
}
printf("%d\n",ans);
return 0;
}
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