SOJ 4107 : consecutive

Description

you are given a 01 sequence whose size is n. you can change at most k 0 into 1 in the sequence.
now I want to know how many consecutive 1 in the sequce at most after you do the change. 

Input

The first of input is an integer t  which stands for the number of test cases.
for each test case, the first line contains two number n, k ( 1 <=  n <= 100000,  0 <= k <= n ).
the next contains n element of the sequence, whose value is 0 or 1.

Output

output the answer in one line for each test case.

Sample Input

2
4 2
1 0 0 1
4 1
1 0 1 0

Sample Output

4
3

Source

baihacker&onmylove@scuacm
Sichuan University Programming Contest 2012 Preliminary

//分析：其中i-j中0的个数可以表达如下：j-i+1-(sum[j]-sum[i-1]);其中sum[i]表示前i个数之和。

分析表达式知假如i是确定的，那么j-i+1和sum[j]-sum[i-1]都是随着j而递增的，所以我们可以从i=1...n，每次遍历i为起点，要找到一个最大的j使得它和i之间的0的个数不大于k，由于之前的表达式是递增的所以可以每次二分出一个j，最后取最优，时间复杂度为O(n*logn)；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define maxn 100005

struct NODE
{
    int ti,num;
}node[maxn];

int a[maxn];
int main()
{
    int t,n,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        int i,j;
        memset(node,0,sizeof(node));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            node[i].ti=i;
            node[i].num+=node[i-1].num+a[i];
        }
        
        int tmp=0,best=0,flag=0;
        for(i=1;i<=n;i++)
        {
            int lif=i,rig=n,mid=(lif+rig)>>1;
            flag=0;
            while(lif<=rig)
            {
                if((node[mid].ti-node[i].ti+1-(node[mid].num-node[i-1].num)>k))
                    rig=mid-1;
                else
                    lif=mid+1;
                mid=(lif+rig)>>1;
            }
            if(mid-i+1>best)
                best=mid-i+1;
        }
        printf("%d\n",best);
    }    
    return 0;
}