ACM学习历程—HDU1003 Max Sum（dp && 最大子序列和）

Description

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

 

这是一道求最大子序列和的题。

思路就是考虑到对于S(i...k) + S(k+1...j) = S(i...j)，如果S(i...k)小于0，自然考虑S(k+1...j)这段和；反之，考虑S(i...j)。

于是从1到n，判断当前的S(i...k)是否小于0，大于0则保留，否则舍去。

考虑到可能整个过程可能S(i...k)一直小于0，所以即使小于0，也要保留当前值now，将其与ans比较。

代码:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <vector>

using namespace std;

int n;
int ans, from, to;

void Work()
{
    from = -1;
    to = -1;
    int k, now, u = -1, v = -1;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d", &k);
        if (u == -1 || now < 0 || now+k < 0)
        {
            u = v = i;
            now = k;
        }
        else
        {
            v = i;
            now = now+k;
        }
        if (from == -1 || now > ans)
        {
            ans = now;
            from = u;
            to = v;
        }
    }
}

int main()
{
    //freopen("test.in", "r", stdin);
    int T;
    scanf("%d", &T);
    for (int times = 1; times <= T; ++times)
    {
        Work();
        if (times != 1)
            printf("\n");
        printf("Case %d:\n", times);
        printf("%d %d %d\n", ans, from, to);
    }
    return 0;
}