HDU-1003 Max Sum (dp)

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本文介绍了一个经典的动态规划问题——寻找整数序列中具有最大和的子序列，并提供了详细的解题思路及C++实现代码。

题目：

Given a sequence a[1],a[2],a[3]……a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output

Case 1:
14 1 4

Case 2:
7 1 6

思路：

经典入门dp，难度小，主要因为自己太粗心WA了4次，记录一下长记性。
WA的地方是记录子序列末位置last时没有初始化为1，导致如果最大子序列就是第一个元素的时候last为0。

Code：

#include<bits/stdc++.h>
using namespace std;
#define MAXN 100000+10
#define M(a, b) memset(a, b, sizeof(a))
int a[MAXN], dp[MAXN], first, last, maxsum;

void Do(){
    M(dp, 0);
    M(a, 0);
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    dp[1] = a[1];
    for (int i = 2; i <= n; ++i) dp[i] = max(a[i]+dp[i-1], a[i]);
    maxsum = dp[1];
    last = 1;
    for (int i = 2; i <= n; ++i)
        if (dp[i] > maxsum){
            maxsum = dp[i];
            last = i;
        }
    int temp = maxsum;
    for (int i = last; i >= 1; --i){
        temp -= a[i];
        if (!temp) first = i;
    }
}

int main()
{
    int T;
    scanf("%d", &T);
    for (int i = 1; i <= T; ++i){
        Do();
        printf("Case %d:\n%d %d %d\n", i, maxsum, first, last);
        if (i != T) printf("\n");
    }

    return 0;
}