HDU -> ACM -> Max Sum

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

 

#include<iostream>
#include<algorithm>
#include<stdio.h>

using namespace std;

int l[1001];
int flag[1001][1001];//当把两个数组放到这里时不会出现栈溢出
struct Mice
{
    int w;
    int s;
    int id;
};
Mice mices[1001];
int cmp(struct Mice a,struct Mice b)
{
    if(a.w!=b.w) return a.w<b.w;
    else return a.s>b.s;
}
int main()
{
    //int l[1001];//存储最大递减序列的长度的值
    //int flag[1001][1001];//存储递减序列下表的值
    int i,size,j,k,max,t;
    i=0;
    while(scanf("%d %d",&mices[i].w,&mices[i].s)!=EOF)
    {
        mices[i].id=i+1;
        i++;
    }
    size=i;
    sort(mices,mices+size,cmp);
    for(i=0;i<size;i++)
    {
        l[i]=1;
        flag[i][0]=i;
    }
    for(i=1;i<size;i++)
    {
        for(j=i-1;j>=0;j--)
        {
            if(mices[i].w>mices[j].w && mices[i].s<mices[j].s && l[i]<l[j]+1)
            {
                max=l[j]+1;
                l[i]=max;

            for(k=0;k<max-1;k++)
                flag[i][k]=flag[j][k];
            flag[i][max-1]=i;
            }
        }
    }
    int index;
        for(index=0,k=1;k<size;k++)
         if(l[index]<l[k]) index=k;

    cout<<l[index];

    for(i=0;i<l[index];i++)
        cout<<endl<<mices[flag[index][i]].id;
    return 0;
}