暑期集训-dp（1）

HDU 1003 MAX SUM  

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

这是早前就训练过的题，所以没什么难度。

列出状态转移方程，用ans[i]记录以data[i]为子列终点的最大值；

ans[i]+=(ans[i-1]>0)?ans[i-1]:0;

dp水题，只要在找出MAX_SUM后找出始末下标就可，注意输出格式。

代码：

#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<iostream>
#include<stack>
#include<queue>
#include<limits>
using namespace std;
#define debug 0
const int maxn=100000+5;
int data[maxn],ans[maxn],caseSum,numberNum,caseNum,maxSum,startPos,endPos;
void Do()
{
    int j;
        for(int i=1;i<=numberNum;i++)
        {
            if(ans[i]>maxSum)
            {
                maxSum=ans[i];
                startPos=endPos=i;
            }
        }
        for(j=endPos-1;ans[j]>=0&&j>0;j--);
        startPos=j+1;
        if(caseNum>1)
        {
            putchar('\n');
        }
        printf("Case %d:\n%d %d %d\n",caseNum,maxSum,startPos,endPos);
}
int main()
{
#if debug
    freopen("in.txt","r",stdin);
#endif // debug
    while(~scanf("%d",&caseSum))
    {
        for(caseNum=1;caseNum<=caseSum;caseNum++)
        {
            maxSum=INT_MIN;
            memset(data,0,sizeof(data));
            scanf("%d",&numberNum);
            for(int i=1;i<=numberNum;i++)
            {
                scanf("%d",&data[i]);
                ans[i]=data[i];
                if(ans[i-1]>0)
                    ans[i]+=ans[i-1];
            }
            Do();
        }
    }
    return 0;
}