越来越水了。。。

这道题是简单的单调队列,同时维护最大值和最小值即可。

另解:multiset大法求区间最大最小,但是复杂度会上升。。。

 

 1 /**************************************************************
 2     Problem: 1342
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:916 ms
 7     Memory:12524 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11  
12 using namespace std;
13 const int N = 1000005;
14 int n, m, C, a[N];
15 int q1[N], q2[N], h1, h2, t1, t2;
16 bool f;
17  
18 inline int read(){
19     int x = 0;
20     char ch = getchar();
21     while (ch < '0' || ch > '9')
22         ch = getchar();
23     while (ch >= '0' && ch <= '9'){
24         x = x * 10 + ch - '0';
25         ch = getchar();
26     }
27     return x;
28 }
29  
30 int pr[10], NUM = 0;
31 inline int print(int x){
32     while (x)
33         pr[++NUM] = (x % 10) + '0', x /= 10;
34     while (NUM)
35         putchar(pr[NUM--]);
36     putchar('\n');
37 }
38  
39 inline bool check(int i){
40     return a[q1[h1]] - a[q2[h2]] <= C && i >= m;
41 }
42  
43 int main(){
44     n = read(), m = read(), C = read();
45     int i;
46     for (i = 1; i <= n; ++i)
47         a[i] = read();
48     f = 0;
49     q1[1] = q2[1] = 1, h1 = t1 = h2 = t2 = 1;
50     if (check(1)){
51         print(1);
52         f = 1;
53     }
54     for (i = 2; i <= n; ++i){
55         while (q1[h1] + m <= i) ++h1;
56         while (h1 <= t1 && a[q1[t1]] <= a[i]) --t1;
57         q1[++t1] = i;
58         while (q2[h2] + m <= i) ++h2;
59         while (h2 <= t2 && a[q2[t2]] >= a[i]) --t2;
60         q2[++t2] = i;
61         if (check(i)){
62             print(i - m + 1);
63             f = 1;
64         }
65     }
66     if (!f) puts("NONE");
67     return 0;
68 }
View Code

(p.s. 那个300ms的这是怎么做到的。。。我输入输出外挂都开了好不好。。。哭T T)

posted on 2014-10-31 23:27  Xs酱~  阅读(238)  评论(3编辑  收藏  举报