动态规划（Dynamic Programming, DP）---- 最大连续子序列和

动态规划（Dynamic Programming, DP）是一种用来解决一类最优化问题的算法思想，简单来使，动态规划是将一个复杂的问题分解成若干个子问题，或者说若干个阶段，下一个阶段通过上一个阶段的结果来推导出，为了避免重复计算，必须把每阶段的计算结果保存下来，方便下次直接使用。

动态规划有递归和递推两种写法。一个问题必须拥有重叠子问题和最优子结构才能使用动态归来来解决，即一定要能写出一个状态转移方程才能使用动态规划来解决。

最大连续子序列和：

令状态dp[i]表示以A[i]作为末尾的连续序列的最大和，

代码：

 // 最大连续子序列和
 #include <stdio.h>
 #include <algorithm>
 using namespace std;

 const int maxn = ;
 int A[maxn], dp[maxn];        // A[i]存放序列，dp[i]存放以A[i]结尾的连续序列的最大和
 int main()
 {
     freopen("in.txt", "r", stdin);
     int n;
     scanf("%d", &n);
     for (int i = ; i < n; i++){
         scanf("%d", &A[i]);
     }

     // 边界
     dp[] = A[];
     for (int i = ; i < n; i++){
         // 状态转移方程
         dp[i] = max(dp[i - 1] + A[i], A[i]);
     }

     // dp[i]存放以A[i]结尾的连续序列的最大值，需要遍历 i 得到最大的才是结果
     int k = ;
     for (int i = ; i < n; i++){
         if (dp[i] > dp[k]){
             k = i;
         }
     }

     printf("%d\n", dp[k]);
     fclose(stdin);
     return ;
 }

题型实战：

　　　　　　　　　　　　　　1007 Maximum Subsequence Sum (25分)

Given a sequence of K integers { N​1​​, N​2​​, ..., N​K​​ }. A continuous subsequence is defined to be { N​i​​, N​i+1​​, ..., N​j​​ } where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (≤10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4

代码：

 #include <stdio.h>
 #include <algorithm>
 using namespace std;

 const int maxn = ;

 int A[maxn];
 // 两个数组
 int firsts[maxn], dp[maxn];

 int main()
 {
     // 读入数据
     // freopen("in.txt", "r", stdin);
     int n;
     scanf("%d", &n);
     for (int i = ; i < n; i++){
         scanf("%d", &A[i]);
     }

     // 边界
     dp[] = A[], firsts[] = ;
     // 如果d[i - 1] > 0, firsts[i] = first[i - 1], 否则firsts[i] = i;
     for (int i = ; i < n; i++){
         if (dp[i - ] >= ){
             firsts[i] = firsts[i - ];
             dp[i] = dp[i - ] + A[i];
         }
         else{
             firsts[i] = i;
             dp[i] = A[i];
         }
     }
     // 遍历dp,寻找最大的子序列和
     int k = ;
     for (int i = ; i < n; i++){
         if (dp[i] > dp[k]){
             k = i;
         }
     }

     // 输出
     if (dp[k] < ){            // 如果最大子序列和都小于0，说明所有元素都是负数
         printf("0 %d %d\n", A[], A[n - ]);
     }
     else{
         printf("%d %d %d\n", dp[k], A[firsts[k]], A[k]);
     }

     // fclose(stdin);

     return ;
 }