poj 2479 Maximum sum（DP）

题意：求两段连续和使之最大

思路：两个方向都处理。先正方向求出到每个index的连续最大和，然后从前往后把每个位置的最大和都更新一下（更新为前边所有和中最大的那个），反方向做同样的处理。

然后i 1..n枚举

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 #include<string>

 6 #include<queue>

 7 #include<algorithm>

 8 #include<map>

 9 #include<iomanip>

10 #include<climits>

11 #include<string.h>

12 #include<numeric>

13 #include<cmath>

14 #include<stdlib.h>

15 #include<vector>

16 #include<stack>

17 #include<set>

18 #define INF 1e7

19 #define MAXN 100010

20 #define maxn 1000010

21 #define Mod 1000007

22 #define N 1299800

23 using namespace std;

24 typedef long long LL;

25 

26 int T;

27 int n, num[50010];

28 int sum1[50010],sum2[50010];

29 

30 void run()

31 {

32     scanf("%d", &n);

33     memset(sum1, 0, sizeof(sum1));

34     memset(sum2, 0, sizeof(sum2));

35     memset(num, 0, sizeof(num));

36     for (int i = 1; i <= n; ++i) 

37         scanf("%d",&num[i]);

38     if (n == 2) {

39         printf("%d\n",num[1] + num[2]);

40         return;

41     }

42     for (int i = 1; i <= n; ++i) {

43         if (sum1[i - 1] > 0)

44             sum1[i] = sum1[i - 1] + num[i];

45         else sum1[i] = num[i];

46     }

47     for (int i = 1; i <= n; ++i)

48         sum1[i] = max(sum1[i - 1], sum1[i]);

49     for (int i = n; i >= 1; --i) {

50         if (sum2[i + 1] > 0)

51             sum2[i] = sum2[i + 1] + num[i];

52         else sum2[i] = num[i];

53     }

54     for (int i = n; i >= 1; --i)

55         sum2[i] = max(sum2[i],sum2[i+1]);

56     int ans = -INF;

57     for (int i = 1; i <= n; ++i)

58         ans = max(ans, sum1[i] + sum2[i + 1]);

59     printf("%d\n",ans);

60 }

61 

62 int main()

63 {

64     scanf("%d", &T);

65     while (T--)

66         run();

67     return 0;

68 }