[CF660C]Hard Process(尺取法)

题目链接：http://codeforces.com/problemset/problem/660/C

尺取法，每次遇到0的时候补一个1，直到补完或者越界为止。之后每次从左向右回收一个0点。记录路径用两个指针卡住，每次更新即可。

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <iomanip>
 4 #include <cstring>
 5 #include <climits>
 6 #include <complex>
 7 #include <fstream>
 8 #include <cassert>
 9 #include <cstdio>
10 #include <bitset>
11 #include <vector>
12 #include <deque>
13 #include <queue>
14 #include <stack>
15 #include <ctime>
16 #include <set>
17 #include <map>
18 #include <cmath>
19 
20 using namespace std;
21 
22 const int maxn = 666666;
23 int n, k;
24 int a[maxn];
25 
26 int main() {
27     // freopen("in", "r", stdin);
28     while(~scanf("%d %d", &n, &k)) {
29         for(int i = 1; i <= n; i++) {
30             scanf("%d", &a[i]);
31         }
32         int ll = 1;
33         int rr = 1;
34         int al = 0;
35         int ar = 0;
36         int cur = 0;
37         int ans = 0;
38         while(1) {
39             if(rr > n) break;
40             while(rr <= n && cur <= k) {
41                 if(a[rr] == 0) {
42                     if(cur == k) break;
43                     cur++;
44                 }
45                 rr++;
46             }
47             if(rr - ll > ans) {
48                 ans = rr - ll;
49                 al = ll;
50                 ar = rr - 1;
51             }
52             while(ll <= rr && a[ll] == 1) ll++;
53             cur--;
54             ll++;
55         }
56         printf("%d\n", ans);
57         for(int i = 1; i <= n; i++) {
58             if(i >= al && i <= ar) printf("1 ");
59             else printf("%d ", a[i]);
60         }
61         printf("\n");
62     }
63     return 0;
64 }

 

posted @ 2016-04-18 18:39  Kirai  阅读(346)  评论(0)  编辑  收藏  举报