HDU - 5773 The All-purpose Zero（思维+LIS）
2017-09-04 10:03:25         来源：K键盘里的青春K

HDU - 5773 The All-purpose Zero（思维+LIS）。

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.

PS:可能举的例子不合适,需要自己好好理解了

```#include
#include
#include
#include
using namespace std;
const int maxn = 1e5 + 5;
int a[maxn], pre[maxn], num[maxn], b[maxn];
int main()
{
int t, n, ca = 1;
scanf("%d", &t);
while(t--)
{
memset(pre, 0, sizeof(pre));
memset(b, 0, sizeof(b));
memset(num, 0, sizeof(num));
scanf("%d", &n);
int cnt = 0, tot = 0;
for(int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
if(a[i] == 0)
pre[i] = pre[i-1] + 1, cnt++;
else
pre[i] = pre[i-1];
}
for(int i = 1; i <= n; i++)
{
if(a[i] != 0)
num[tot++] = a[i] - pre[i];
}
int len = 1, l , mid, r;
if(tot == 0) len = 0;
b[1] = num[0];
for(int i = 1; i < tot; i++)
{
l = 1, r = len;
while(l <= r)
{
mid = (l+r)/2;
if(b[mid] < num[i]) l = mid+1;
else r = mid - 1;
}
b[l] = num[i];
if(l > len) len++;
}
printf("Case #%d: %d\n", ca++, len+cnt);
}
return 0;
}
```