Max Sum（经典DP）

Max Sum Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1003

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

 

设一数组sum，sum[i]含义：所有以a[i]为结尾的序列的序列和构成一个集合，此集合的最大值就是sum[i]
如1,2,3.所有以a[2]=3为结尾的序列的序列和集合是{6,5,3},因而sum[2]=6.

sum的状态转移方程：sum[i] = max{sum[i-1]+a[i], a[i]}
ans必定是sum[0···(k-1)]之一。由于要记录起始位置和结束位置，引入s数组记录获得sum的序列的起始元素的位置，而由sum的定义，sum[i]的结束位置是i不用另外记录。

 

//Memory: 1420 KB         Time: 0 MS
//Language: C++         Result: Accepted
#include <iostream>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
#define cti const int
#define ctll const long long
#define dg(i) cout << "*" << i << endl;

int a[100005], sum[100005], s[100005];

int main()
{
    int T, N;
    int ca = 0, ans;
    scanf("%d", &T);
    while(ca < T)
    {
        scanf("%d", &N);
        for(int i = 0; i < N; i++)
            scanf("%d", &a[i]);
        ans = 0;
        sum[0] = a[0];
        s[0] = 0;
        for(int i = 1; i < N; i++)
        {
            if(sum[i-1] >= 0)
            {
                sum[i] = sum[i-1] + a[i];
                s[i] = s[i-1];
            }
            else
            {
                sum[i] = a[i];
                s[i] = i;
            }
            if(sum[ans] < sum[i]) ans = i;
        }
        printf("Case %d:\n%d %d %d\n", ++ca, sum[ans], s[ans] + 1, ans + 1);
        if(ca != T) puts("");
    }
    return 0;
}