Almost Sorted Array
Time Limit: 4000/2000 MS (Java/Others)????Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 10562????Accepted Submission(s): 2449
Problem Description
We are all familiar with sorting algorithms: quick sort,merge sort,heap sort,insertion sort,selection sort,bubble sort,etc. But sometimes it is an overkill to use these algorithms for an almost sorted array.
We say an array is sorted if its elements are in non-decreasing order or non-increasing order. We say an array is almost sorted if we can remove exactly one element from it,and the remaining array is sorted. Now you are given an array?
a1,a2,…,an,is it almost sorted?
?
?
Input
The first line contains an integer?
T?indicating the total number of test cases. Each test case starts with an integer?n?in one line,then one line with?n?integers?a1,a2,…,an.
1≤T≤2000
2≤n≤105
1≤ai≤105
There are at most 20 test cases with?n>1000.
?
Output
For each test case,please output "`YES`" if it is almost sorted. Otherwise,output "`NO`" (both without quotes).
?
?
Sample Input
3
3
2 1 7
3
3 2 1
5
3 1 4 1 5
?
?
Sample Output
?
?
Source
2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）
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?
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遍历数组，如果a >= dp[len]，接到后面，否则，在dp中寻找第一个大于这a的数，把他替换掉，原因很好想，要想使得序列足够长，那么dp[i]越小越好。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100000 + 5;
int n,val[maxn];
int cnt1,cnt2,dp[maxn];
int main() {
int t;
int a,temp;
scanf("%d",&t);
while(t --) {
cnt1 = 0,cnt2 = 0;
memset(dp,0,sizeof dp);
scanf("%d",&n);
for(int i = 0; i < n; i ++) scanf("%d",&val[i]);
for(int i = 0; i < n; i ++) {
if(val[i] >= dp[cnt1]) dp[++ cnt1] = val[i];
else {
temp = upper_bound(dp + 1,dp + 1 + cnt1,val[i]) - dp;
dp[temp] = val[i];
}
}
memset(dp,sizeof dp);
for(int i = n - 1; i >= 0; i --) {
if(val[i] >= dp[cnt2]) dp[++ cnt2] = val[i];
else {
temp = upper_bound(dp + 1,dp + 1 + cnt2,val[i]) - dp;
dp[temp] = val[i];
}
}
if(cnt1 >= n - 1 || cnt2 >= n - 1) printf("YESn");
else printf("NOn");
}
return 0;
}
