HDU 1003 Max Sum - from lanshui_Yang

Problem 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

题目大意：给你n个数s1，s2，s3……,sn，求最大子串和，即求max{si + …… + sj }（1 <= i <= j <= n )。

解题思路:这道题基础动态规划的典型问题，设f[i] 表示数列s0 , s1 ，…… , si 中 包含s[i] 的最大子串和，则由状态f[i - 1] 转移到 状态f[i] 共有两种决策：

1、当f[i - 1] >= 0 时 ，f[i] = f[i - 1] + s[i] ; 

2、当f[i - 1] < 0 时 ，f[i] = s[i]。

由于本题要记录起点和终点 ， 所以需要一个标记数组p ， p[i] == true表示数列s0 , s1 ，…… , si 中 包含s[i] 的最大子串和 由 f[i - 1] 和 s[i] 共同构成；否则，则表示只由s[i] 构成。

具体请看代码：

#include
#include
#include
#include
#include
#include
using namespace std ;
const int MAXN = 1e6 + 5 ;
int s[MAXN] ;
int f[MAXN] ;
bool p[MAXN] ;
int start , end ;
int MAX ;
int n ;
void init()
{
    scanf("%d" , &n) ;
    int i ;
    for(i = 0 ; i < n ; i ++)
    {
        scanf("%d" , &s[i]) ;
    }
}
void solve()
{
    MAX = s[0] ;
    end = 0 ;
    p[0] = false ;
    f[0] = s[0] ;
    int i ;
    for(i = 1 ; i < n ; i ++)
    {
        if(f[i - 1] >= 0) // 注意：题目中要求输出第一个符合条件的起始点，所以是 " >= " 
        {
            f[i] = f[i - 1] + s[i] ;
            p[i] = true ;
        }
        else
        {
            f[i] = s[i] ;
            p[i] = false ;
        }
        if(MAX < f[i])
        {
            MAX = f[i] ;
            end = i ;
        }
    }
}
int main()
{
    int T ;
    scanf("%d" , &T) ;
    int ca = 0 ;
    int first = true ;
    while (T --)
    {
        init() ;
        solve() ;
        int i ;
        for(i = end ; i >= 0 ; i --)
        {
            if(p[i] == false)
            {
                start = i ;
                break ;
            }
        }
        if(first)
        {
            first = false ;
        }
        else
        puts("") ;
        printf("Case %d:\n" , ++ ca) ;
        printf("%d %d %d\n" , MAX , start + 1 , end + 1) ;
    }
    return 0 ;
}