【hdu1024】简单dp

http://acm.hdu.edu.cn/showproblem.php?pid=1024 最大m字段和,题目就不多说了,经典dp

这题坑爹。。。首先不说明m的范围(n<=1000000),还以为有O(n)的算法,吓得不敢做。其次,按照题目给的范围完全可能超int,但是数据事实上int远远够了,害得我__int64读比读int慢两倍,直接TLE。就是因为TLE的原因,害得我数组初始化没有考虑全(其实应该怪自己没有考虑清楚->_->),但事实上,完全可以每次memset一遍,因为超时,不敢==。各种原因,几分钟敲完的水题调了1个多小时!我也是醉了。。。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <map>
 7 #include <stack>
 8 #include <string>
 9 #include <ctime>
10 #include <queue>
11 #define mem0(a) memset(a, 0, sizeof(a))
12 #define mem(a, b) memset(a, b, sizeof(a))
13 #define lson l, m, rt << 1
14 #define rson m + 1, r, rt << 1 | 1
15 #define eps 0.0000001
16 #define lowbit(x) ((x) & -(x))
17 #define memc(a, b) memcpy(a, b, sizeof(b))
18 #define x_x(a) ((a) * (a))
19 #define LL __int64
20 #define DB double
21 #define pi 3.14159265359
22 #define MD 10000007
23 #define INF (int)1e9
24 #define max(a, b) ((a) > (b)? (a) : (b))
25 using namespace std;
26 int f[1000010][2][2];
27 int a[1000010];
28 int main()
29 {
30         //freopen("input.txt", "r", stdin);
31         //freopen("output.txt", "w", stdout);
32         int n, m;
33         while(~scanf("%d%d", &m, &n))  {
34                 for(int i = 1; i <= n; i++) {
35                         scanf("%d", a + i);
36                 }
37                 for(int i = 0; i <= n; i++) {
38                         f[i][0][0] = 0;
39                         f[i][0][1] = -INF;
40                 }
41                 int now = 1;
42                 for(int j = 1; j <= m; j++) {
43                         f[0][now][0] = f[0][now][1] = - INF;
44                         for(int i = 1; i <= n; i++) {
45                                 if(i < j) {
46                                         f[i][now][1] = f[i][now][0]= -INF;
47                                         continue;
48                                 }
49                                 f[i][now][1] = max(max(f[i - 1][now ^ 1][0], f[i - 1][now ^ 1][1]), f[i - 1][now][1]) + a[i];
50                                 f[i][now][0] = max(f[i - 1][now][0], f[i - 1][now][1]);
51                         }
52                         now ^= 1;
53                 }
54                 printf("%d
", max(f[n][now ^ 1][0], f[n][now ^ 1][1]));
55         }
56         return 0;
57 }
View Code
原文地址:https://www.cnblogs.com/jklongint/p/4077745.html