HDOJ-----5773The All-purpose Zero(LIS)

最新推荐文章于 2017-09-07 15:44:30 发布

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最新推荐文章于 2017-09-07 15:44:30 发布

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分类专栏： LIS/LCS

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本文介绍了一种解决最长递增子序列问题的算法实现，特别关注如何处理序列中可变元素（如0可视为任意值）的情况。通过动态规划方法，文章详细解释了如何寻找给定整数序列中最长的严格递增子序列。

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The All-purpose Zero

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1621 Accepted Submission(s): 775

Problem Description

?? gets an sequence S with n intergers(0 < n <= 100000,0<= S[i] <= 1000000).?? has a magic so that he can change 0 to any interger(He does not need to change all 0 to the same interger).?? wants you to help him to find out the length of the longest increasing (strictly) subsequence he can get.

Input

The first line contains an interger T,denoting the number of the test cases.(T <= 10)
For each case,the first line contains an interger n,which is the length of the array s.
The next line contains n intergers separated by a single space, denote each number in S.

Output

For each test case, output one line containing “Case #x: y”(without quotes), where x is the test case number(starting from 1) and y is the length of the longest increasing subsequence he can get.

Sample Input

2
7
2 0 2 1 2 0 5
6
1 2 3 3 0 0

Sample Output

Case #1: 5
Case #2: 5

Hint
In the first case,you can change the second 0 to 3.So the longest increasing subsequence is 0 1 2 3 5.

很明显的LIS，可以把0变为任意数字，求最长递增子序列

这里有个坑就是0可以变为负值，只要对0做一些特殊处理就可以了

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 100010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int dp[maxn], cnt[maxn];
int ans, num, n;
void lis(){
    int flag;
    ans = 1;
    num = 0;
    dp[0] = -inf;
    dp[1] = cnt[1];
    for(int i = 2; i <= n; i++){
        if(dp[ans] < cnt[i]){
            dp[++ans] = cnt[i];
        }
        else if(!cnt[i]){
            for(int j = ans; j >= num; j--){
                dp[j+1] = dp[j]+1;//dp[0]是不变的，所以每次cnt[i]为0时把前边的dp[j]的lis向后推1位并都加上1就可以动态更新dp[i]的lis
            }
            ans++;
            num++;
        }
        else{
            flag = lower_bound(dp, dp+ans, cnt[i])-dp;
            dp[flag] = cnt[i];
        }
    }
}
int main(){
	int t, kcase = 1;
	scanf("%d", &t);
	while(t--){
        scanf("%d", &n);
        for(int i = 1; i <= n; i++){
            scanf("%d", &cnt[i]);
        }
        lis();
        printf("Case #%d: %d\n", kcase++, ans);
	}
	return 0;
}